The proof is constructive. We set
\(k_1:=\Big (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}}\Big )^T,\quad k_2:=\Big (-\frac{1}{2},\frac{2}{\sqrt{6}}\Big )^T,\quad k_3:=\Big (\frac{1}{2},0\Big )^T.\) 

A straightforward computation gives
\(k_1\otimes k_1=\left(\begin{array}{cc}\frac{1}{2} & \frac{1}{\sqrt{6}}\\\frac{1}{\sqrt{6}} & \frac{1}{3}\end{array}\right),\qquad k_2\otimes k_2=\left(\begin{array}{cc}\frac{1}{4} & -\frac{1}{\sqrt{6}}\\-\frac{1}{\sqrt{6}} & \frac{2}{3}\end{array}\right),\qquad k_3\otimes k_3=\left(\begin{array}{cc}\frac{1}{4} & 0\\0 & 0\end{array}\right).\nonumber \) 

It's obvious that \(k_1\otimes k_1, k_2\otimes k_2, k_3\otimes k_3\)  are linearly independent, hence form a basic for \(\mathcal {S}^{2\times 2}\)  and
\(\sum _{i=1}^3k_i\otimes k_i=Id.\) 
Thus, taking \(r_0> 0\)  small, for any symmetric matrix \(R\in B_{r_0}(Id)\)  , the following equation
\(\sum _{i=1}^3\gamma ^2_{k_i}k_i\otimes k_i=R\) 
has unique, positive solution \(\gamma _{k_i}\)  and
\(\Vert \gamma _{k_i}-1\Vert _0\le \frac{1}{2}.\) 
Since the representation is unique, the dependence of \(\gamma _{k_i}\)  on \(R\)  is smooth.
Then, for any \(f\in R^{2}\) , we set
\(\left(\begin{array}{cc}g_{k_1}(f)\\[12pt]g_{k_2}(f)\end{array}\right):=(k_1, k_2)^{-1}f,\) 
where \((k_1, k_2)^{-1}\)  denote the inverse matrix of \((k_1, k_2).\) 
Thus, \(g_{k_1}, g_{k_2}\)  are linear functions and it's obvious that
\(f=\sum _{i=1}^2g_{k_i}(f)k_i, \quad \forall f\in R^2.\) 

Thus, we finished the proof of this lemma.
MM\(\square \)
As in [1]}, we introduce the following operators
in order to deal with the stresses.
The operator \(\mathcal {R}\)  
Suppose that the vector function \(\vec{U}(x)=(U_1(x),U_2(x))^T\in C_c^{\infty }(B_r; C^2)\)  satisfy
\(\int _{R^2}U_i(x)dx=0,\quad \int _{R^2}(x_iU_j-x_jU_i)(x)dx=0,\quad i,j=1,2\) 
 and take the following form
\(\vec{U}(x)=\left(\begin{array}{c}V_1(x)e^{i\lambda k\cdot x}\\V_2(x)e^{i\lambda k\cdot x}\end{array}\right),\nonumber \) 

where \(\lambda >0,\quad k=(k_1, k_2)\in R^2\setminus \lbrace 0\rbrace \) .
We denote \(\mathcal {R}U_o(x)\)  by
\(\mathcal {R}U_o(x):=\left(\begin{array}{cc}\frac{M_{11}(x)}{i\lambda }e^{i\lambda k\cdot x} & \frac{M_{12}(x)}{i\lambda }e^{i\lambda k\cdot x}\\\frac{M_{12}(x)}{i\lambda }e^{i\lambda k\cdot x} & \frac{M_{22}(x)}{i\lambda }e^{i\lambda k\cdot x}\end{array}\right),\nonumber \) 

where \(\vec{M}=(M_{11}, M_{12}, M_{22})\)  satisfy
\(M_{11}k_1+M_{12}k_2=V_1, \quad M_{12}k_1+M_{22}k_2=V_2.\) 

Obviously, the linear equation (REF ) always have a solution
\((M_{11}(x),~M_{12}(x),~M_{22}(x))\in C_c^{\infty }(B_{r},C^3),\quad \Vert M_{ij}\Vert _0\le C_0\Vert \vec{V}\Vert _0.\) 

Here and subsequent in this section, \(C_0\)  denotes a absolute constant.
In fact, we may choose \(M_{12}(x)=0\)  first, if both \(k_1\)  and \(k_2 \)  are not zero, then (REF ) gives \((M_{11}, M_{22})\)  and they satisfy (REF ). In the case one of \(k_1\) , \(k_2\)  is zero, for example \(k_1=0\) , (REF ) gives \((M_{12}, M_{22})\)  and we set \(M_{11}=0\) . They also satisfy (REF ).
We compute
\({\rm div}\mathcal {R}U_o(x)=&\left(\begin{array}{c}V_1(x)e^{i\lambda k\cdot x}\\V_2(x)e^{i\lambda k\cdot x}\end{array}\right)+\left(\begin{array}{c}\frac{\partial _1M_{11}(x)+\partial _2M_{12}(x)}{i\lambda }e^{i\lambda k\cdot x} \\\frac{\partial _1M_{12}(x)+\partial _2M_{22}(x)}{i\lambda }e^{i\lambda k\cdot x}\end{array}\right).\nonumber \) 

Repeat the above process, there exists \((N_{11}(x),~N_{12}(x),~N_{22}(x))\in C_c^{\infty }(B_{r},C^3)\)  such that if we set
\(\mathcal {R}U_{c1}(x):=\left(\begin{array}{cc}\frac{N_{11}(x)}{(i\lambda )^2}e^{i\lambda k\cdot x} & \frac{N_{12}(x)}{(i\lambda )^2}e^{i\lambda k\cdot x}\\\frac{N_{12}(x)}{(i\lambda )^2}e^{i\lambda k\cdot x} & \frac{N_{22}(x)}{(i\lambda )^2}e^{i\lambda k\cdot x}\end{array}\right),\nonumber \) 

then
\({\rm div}\mathcal {R}U_{c1}(x)=-\left(\begin{array}{c}\frac{\partial _1M_{11}(x)+\partial _2M_{12}(x)}{i\lambda }e^{i\lambda k\cdot x} \\\frac{\partial _1M_{12}(x)+\partial _2M_{22}(x)}{i\lambda }e^{i\lambda k\cdot x}\end{array}\right)+\left(\begin{array}{c}\frac{\partial _1N_{11}(x)+\partial _2N_{12}(x)}{(i\lambda )^2}e^{i\lambda k\cdot x} \\\frac{\partial _1N_{12}(x)+\partial _2N_{22}(x)}{(i\lambda )^2}e^{i\lambda k\cdot x}\end{array}\right)\nonumber \) 

and
\(\Vert N_{ij}\Vert _0\le C_0\Vert \nabla M\Vert _0\le C_0\Vert \nabla V\Vert _0,\quad \Vert N_{ij}\Vert _1\le C_0\Vert \nabla ^2 V\Vert _0.\nonumber \) 

A straightforward computation gives
\({\rm div}\Big (\mathcal {R}U_o(x)+\mathcal {R}U_{c1}(x)\Big )&=\left(\begin{array}{c}V_1(x)e^{i\lambda k\cdot x}\\V_2(x)e^{i\lambda k\cdot x}\end{array}\right)+\left(\begin{array}{c}\frac{\partial _1N_{11}(x)+\partial _2N_{12}(x)}{(i\lambda )^2}e^{i\lambda k\cdot x} \\\frac{\partial _1N_{12}(x)+\partial _2N_{22}(x)}{(i\lambda )^2}e^{i\lambda k\cdot x}\end{array}\right).\nonumber \) 

Performing the above process, for any integer \(m\ge 2\) , we have symmetric matrix functions \(\mathcal {R}U_{ci}\in C_c^{\infty }(B_r): i=1,2,\cdot \cdot \cdot ,m-1\)  such that
\(\Vert \mathcal {R}U_{ci}\Vert _0\le C_0\frac{\Vert \nabla ^iV\Vert _0}{\lambda ^{i+1}}\nonumber \) 

and
\({\rm div}\Big (\mathcal {R}U_o(x)+\sum _{i=1}^{m-1}\mathcal {R}U_{ci}(x)\Big )&=\left(\begin{array}{c}V_1(x)e^{i\lambda k\cdot x}\\V_2(x)e^{i\lambda k\cdot x}\end{array}\right)+\left(\begin{array}{c}\frac{R_1}{(i\lambda )^m}e^{i\lambda k\cdot x} \\\frac{R_2}{(i\lambda )^m}e^{i\lambda k\cdot x}\end{array}\right)\nonumber \) 

with
\(\Vert R_1\Vert _0+\Vert R_2\Vert _0\le C_0\Vert \nabla ^mV\Vert _0.\) 
Since \(\mathcal {R}U_o(x)+\sum \limits _{i=1}^{m-1}\mathcal {R}U_{ci}(x)\in C_c^{\infty }(B_r)\)  is a symmetric matrix, then
\(\int _{R^2}{\rm div}\Big (\mathcal {R}U_o(x)+\sum _{i=1}^{m-1}\mathcal {R}U_{ci}(x)\Big )=0, \quad \int _{R^2}(x_iH_j-x_jH_i)(x)dx=0,\quad i,j=1,2.\) 

Here we used the notaion \({\rm div}\Big (\mathcal {R}U_o(x)+\sum \limits _{i=1}^{m-1}\mathcal {R}U_{ci}(x)\Big )=(H_1,H_2)^T\) .
By (REF ) and (REF ), if we set
\((K_1, K_2)^T:=\Big (\frac{R_1}{(i\lambda )^m}e^{i\lambda k\cdot x}, \frac{R_2}{(i\lambda )^m}e^{i\lambda k\cdot x}\Big )^T,\) 
we also have
\(K_i\in C_c^{\infty }(B_{r};C),\quad \int _{R^2}K_i(x)dx=0, \quad \int _{R^2}(x_iK_j-x_jK_i)(x)dx=0,\quad i,j=1,2.\nonumber \) 

Together with a result given in [1]}, we know that there exists a symmetric matric function \(\delta \mathcal {R}U\in C_c^{\infty }(B_{r})\)  such that
\({\rm div}\delta \mathcal {R}U=-(K_1,K_2)^T,\quad \Vert \delta \mathcal {R}U\Vert _0\le C_0(r)\Vert K\Vert _0\le C_0(r)\frac{\Vert \nabla ^mV\Vert _0}{\lambda ^m}.\nonumber \) 

Finally, we set \(\mathcal {R}U:=\mathcal {R}U_0+\sum \limits _{i=1}^{m-1}\mathcal {R}U_{ci}(x)+\delta \mathcal {R}U\) , then \(\mathcal {R}U\)  is a symmetric matric function and satisfies
\(\mathcal {R}U\in C_c^{\infty }(B_{r}),\quad {\rm div}\mathcal {R}U=\vec{U}.\nonumber \) 

Moreover, there holds \(\Vert \mathcal {R}U\Vert _0\le C_0(r)\Big (\sum \limits _{i=0}^{m-1}\frac{\Vert V\Vert _i}{\lambda ^{i+1}}+\frac{\Vert V\Vert _m}{\lambda ^m}\Big )\) .
In fact,
\(\Vert \mathcal {R}U_0\Vert _0\le C_0\frac{\Vert V\Vert _0}{\lambda },\quad \Vert \mathcal {R}U_{ci}\Vert _0\le C_0\frac{\Vert V\Vert _i}{\lambda ^{i+1}},i=1,\cdot \cdot \cdot ,m-1,\quad \Vert \delta \mathcal {R}U\Vert _0\le C_0(r)\frac{\Vert V\Vert _m}{\lambda ^m}.\nonumber \) 

Sum them is what we claimed.
Now we introduce a vector space. Put
\(\Xi :=&\Big \lbrace \vec{U}(x):\vec{U}(x)=(U_1(x),U_2(x))^T\in C_c^{\infty }(B_r; C^2),\int _{R^2}U_i(x)dx=0,\nonumber \\&\int _{R^2}(x_iU_j-x_jU_i)(x)dx=0,\quad i,j=1,2 ~~{\quad \rm and \quad }~~\vec{U}(x)=\sum _{j=0}^7\left(\begin{array}{c}U_{1j}(x)e^{i\lambda _j k\cdot x}\\U_{2j}(x)e^{i\lambda _j k\cdot x}\end{array}\right)\Big \rbrace ,\nonumber \) 

where \( k\in R^2\setminus \lbrace 0\rbrace \)  and \(\lambda _j>0, j=0,\cdot \cdot \cdot ,7\) .
Proposition 3.1 
There exists a linear operator \(\mathcal {R}\)  from \(\Xi \)  to \(C_c^{\infty }(B_r; \mathcal {S}^{2\times 2})\)  such that for any \(\vec{U}(x)\in \Xi \)  with
\(\vec{U}(x)=\sum _{j=0}^7\vec{U}_j(x):=\sum _{j=0}^7\left(\begin{array}{c}U_{1j}(x)e^{i\lambda _j k\cdot x}\\U_{2j}(x)e^{i\lambda _j k\cdot x}\end{array}\right),\) 
there holds
\({\rm div}\mathcal {R}U(x)=\vec{U}(x),\quad \Vert \mathcal {R}U\Vert _0\le C_0(r)\sum _{j=0}^7\Big (\sum _{i=0}^{m-1}\frac{\Vert U_{1j}\Vert _i+\Vert U_{2j}\Vert _i}{\lambda _j^{i+1}}+\frac{\Vert U_{1j}\Vert _m+\Vert U_{2j}\Vert _m}{\lambda _j^m}\Big ).\) 
We have defined the operator \(\mathcal {R}\)  on function \(\vec{U}_j(x)=\left(\begin{array}{c}U_{1j}(x)e^{i\lambda _j k\cdot x}\\U_{2j}(x)e^{i\lambda _j k\cdot x}\end{array}\right)\) 
and
\(\Vert \mathcal {R}U_j\Vert _0\le C_0(r)\Big (\sum _{i=0}^{m-1}\frac{\Vert U_{1j}\Vert _i+\Vert U_{2j}\Vert _i}{\lambda _j^{i+1}}+\frac{\Vert U_{1j}\Vert _m+\Vert U_{2j}\Vert _m}{\lambda _j^m}\Big ).\) 
Then, set
\(\mathcal {R}U:=\sum _{j=0}^7\mathcal {R}U_j.\nonumber \) 

It is obvious that \(\mathcal {R}U \in C_c^{\infty }(B_r; \mathcal {S}^{2\times 2})\)  and satisfies (REF ).
MM\(\square \)
The operator \(\mathcal {G}\)  
Let \(f(x)=\varphi (x)e^{i\lambda k\cdot x}\)  with \(\varphi (x)\in C_c^{\infty }(B_r;C)\)  and \(\int _{R^2}\varphi (x)e^{i\lambda k\cdot x}dx=0\) , where \(\lambda > 0\)  and \(k\in R^2\setminus \lbrace 0\rbrace \) . Set
\(\mathcal {G}f_o:=\frac{k}{i\lambda |k|^2}\varphi (x)e^{i\lambda k\cdot x},\) 
then
\({\rm div}\mathcal {G}f_o=f+\frac{k\cdot \nabla \varphi }{i\lambda |k|^2}e^{i\lambda k\cdot x}.\) 
Set
\(\mathcal {G}f_{c1}:=-\frac{k}{(i\lambda )^2|k|^4}k\cdot \nabla \varphi e^{i\lambda k\cdot x},\) 
then
\({\rm div}(\mathcal {G}f_0+\mathcal {G}f_{c1})=f-\frac{(k\cdot \nabla )^2 \varphi }{(i\lambda )^2|k|^4}e^{i\lambda k\cdot x}.\) 
It's obvious that
\(\Vert \mathcal {G}f_o\Vert _0\le C_0\frac{\Vert \varphi \Vert _0}{\lambda },\Vert \mathcal {G}f_{c1}\Vert _0\le C_0\frac{\Vert \varphi \Vert _1}{\lambda ^2}.\) 
Performing this process, for any integer \(m\ge 2\) , there exist \(\mathcal {G}f_{ci}:i=1,2,\cdot \cdot \cdot ,m-1\)  such that
\({\rm div}(\mathcal {G}f_0+\sum _{i=1}^{m-1}\mathcal {G}f_{ci})=f+(-1)^{m-1}\frac{(k\cdot \nabla )^m \varphi }{(i\lambda )^m|k|^{2m}}e^{i\lambda k\cdot x}\) 
and
\(\Vert \mathcal {G}f_{ci}\Vert _0\le C_0\frac{\Vert \varphi \Vert _i}{\lambda ^{i+1}}.\) 
Since
\(\int _{R^2}f(x)dx=0, \quad \int _{R^2}{\rm div}\Big (\mathcal {G}f_0+\sum _{i=0}^{m-1}\mathcal {G}f_{ci}\Big )dx=0,\nonumber \) 

therefore
\(\int _{R^2}\frac{(k\cdot \nabla )^m\varphi }{(i\lambda |k|^2)^m}e^{i\lambda k\cdot x}dx=0.\nonumber \) 

By a result in [3]}, see Chapter 1, we know that there exists \(\mathcal {G}f_{cm}\in C_c^{\infty }(B_r; C^2)\)  such that
\({\rm div}\mathcal {G}f_{cm}=(-1)^{m}\frac{(k\cdot \nabla )^m \varphi }{(i\lambda )^m|k|^{2m}}e^{i\lambda k\cdot x},\quad \Vert \mathcal {G}f_{cm}\Vert _0\le C_0(r)\frac{\Vert \nabla ^m\varphi \Vert _0}{\lambda ^m}.\) 
In fact, there exists \(\mathcal {G}f_{cm}\)  such that
\(\Vert \nabla \mathcal {G}f_{cm}\Vert _4\le C_0(r)\Big \Vert \frac{(k\cdot \nabla )^m \varphi }{(i\lambda )^m|k|^{2m}}\Big \Vert _4\le C_0(r)\frac{\Vert \nabla ^m\varphi \Vert _0}{\lambda ^m}.\nonumber \) 

Moreover, due to \(\mathcal {G}f_{cm}\in C_c^{\infty }(B_r)\) , we have \(\Vert \mathcal {G}f_{cm}\Vert _{W^{1,4}}\le C_0(r)\frac{\Vert \nabla ^m\varphi \Vert _0}{\lambda ^m}\) . Sobolev embedding gives
\(\Vert \mathcal {G}f_{cm}\Vert _0\le C_0(r)\frac{\Vert \nabla ^m\varphi \Vert _0}{\lambda ^m}\)  which is what we claimed.
Finally, we set
\(\mathcal {G}f:=\mathcal {G}f_{0}+\sum _{i=1}^m\mathcal {G}f_{ci},\) 
then
\({\rm div}\mathcal {G}f=f\nonumber \) 

and
\(\mathcal {G}f\in C_c^{\infty }(B_r;C^2),\quad \Vert \mathcal {G}f\Vert _0\le C_0(r)\Big (\sum _{i=0}^{m-1}\frac{\Vert \varphi \Vert _i}{\lambda ^{i+1}}+\frac{\Vert \varphi \Vert _m}{\lambda ^{m}}\Big ).\nonumber \) 

In conclusion, we have
Proposition 3.2 
Let the vector space \(\Psi \)  given by
\(\Psi :=\Big \lbrace H(x): H(x)=\sum _{j=0}^mH_j(x):=\sum _{j=0}^mb_j(x)e^{i\lambda _j k\cdot x},\quad b_j\in C_c^{\infty }(B_{r};C)\quad {\rm and}\quad \int _{R^2}H_j(x)dx=0\Big \rbrace ,\nonumber \) 

then there exists a linear operator \(\mathcal {G}: \Psi \rightarrow C_c^{\infty }(B_{r};C^2)\)  such that for any positive integer \(m\)  and any \(H(x)=\sum \limits _{j=0}^7b_j(x)e^{i\lambda k\cdot x}\in \Psi \) , there holds
\({\rm div}\mathcal {G}(H)(x)=H(x),\quad \Vert \mathcal {G}(H)\Vert _0\le C_0(r)\sum _{j=0}^7\sum _{i=0}^{m-1} \Big (\frac{\Vert b_j\Vert _i}{\lambda _j^{i+1}}+\frac{\Vert b_i\Vert _m}{\lambda _j^m}\Big ).\) 

The proof is similar to that of Proposition REF , we omit it here.
MM\(\square \)
The construction of approximate solutions
The construction of \(\tilde{v},~\tilde{p},~\tilde{\theta },~\tilde{R},~\tilde{g}\)  from \(v,~p,~\theta ,~R,~g\)  consists of several steps. The main idea is to decompose the stress errors into some blocks with the add of geometric lemma and remove one block by constructing new approximate solutions in each step. In this section, we perform the first step.
For convenience, we set \(v_0:=v,~p_0:=p,~\theta _0:=\theta ,~R_0:=R,~g_0:=g\)  and in this section \(C_0\)  denotes a absolute constant.
Partition of unity and Conditions on the parameters
We first introduce a partition of unity. Following the construction given in [4]}, we have the following partition of unity. For two constants \(c_{1}\)  and \(c_{2}\)  such that \(\frac{\sqrt{3}}{2}<c_{1}<c_{2}<1\)  , we have a family of functions \(\alpha _l\in C_c^{\infty }(R^3), l\in Z^3\)  such that
\(\sum \limits _{l\in Z^3}\alpha _l^2=1,\quad \hbox{supp}\alpha _l\subseteq B_{c_2}(l).\) 

Our construction depends on three parameters: \(\ell , \mu _1, \lambda _1\)  and we assume they satisfy the following inequalities:
\(\mu _1\ge \frac{\Lambda }{\delta }\ge 1,\quad \ell ^{-1}\ge \frac{\Lambda }{\eta \delta }\ge 1,\quad \lambda _1\ge \max \lbrace \mu _1^{1+\varepsilon },\ell ^{-(1+\varepsilon )}\rbrace .\) 

Decomposition of stress error
First, we apply Geometric Lemma REF  to obtain \(r_{0}>0\)  and vectors
\(k_1=\Big (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}}\Big )^T,\quad k_2=\Big (-\frac{1}{2},\frac{2}{\sqrt{6}}\Big )^T,\quad k_3=\Big (\frac{1}{2},0\Big )^T\nonumber \) 

which are given in the proof of lemma REF , see (REF ),
together with corresponding functions
\(\gamma _{k_i}\in C^{\infty }(B_{r_{0}}(Id)),\quad g_{k_i}\in C^{\infty }(R^{2}),\quad i=1,2,3,\) 

where \(g_{k_3}=0\) .
Next, we let \(\varphi \in C_c^\infty (R^2\times R)\)  be a standard nonnegative radial function and denote the corresponding family of mollifiers by
\(\varphi _{\ell }(t,x):=\frac{1}{\ell ^3}\varphi \Big (\frac{t}{\ell },\frac{x}{\ell }\Big ).\nonumber \) 

Then set
\(f_{0\ell }(t,x):=f_0\ast \varphi _\ell (t,x),\qquad R_{0\ell }(t,x):=R_0\ast \varphi _\ell (t,x).\nonumber \) 

By Lemma REF , we decompose \(f_{0}\)  as
\(f_{0}(t,x)=\sum \limits _{i=1}^{2}g_{k_i}(f_{0})(t,x)k_i:=-\sum \limits _{i=1}^{2}c_i(t,x)k_i.\nonumber \) 

Here we denote \(c_i(t,x):=-g_{k_i}(f_{0})(t,x)\) .
Thus
\(f_{0\ell }(t,x)=-\sum \limits _{i=1}^{2}c_{i\ell }(t,x)k_i.\) 

Since \(g_{k_1}\)  is linear function,
\(c_{i\ell }(t,x)=-g_{k_i}(f_{0\ell })(t,x).\) 

By (), we know that
\(c_{i\ell }(t,x)\in C^{\infty }_c(Q_{r+\ell }),~\Vert c_i\Vert _0\le 2\delta .\) 

Then, we introduce \(\rho (t,x)\)  as following
\(\rho (t,x)\in C_c^\infty (Q_{r+\delta }),\quad \rho (t,x)=\sqrt{2\delta }~~~ {\rm in}~~~ Q_{r+\frac{\delta }{2}},\quad 0\le \rho (t,x)\le \sqrt{2\delta },\quad \Vert \rho \Vert _{C^k_{t,x}}\le C_0\delta ^{-\frac{1}{2}-(k-1)}\) 

and set
\(e(t,x):=\rho ^2(t,x).\) 

By (REF ), parameter assumption (REF ), (REF ) and (REF ), we have
\(\Big \Vert \frac{R_{0\ell }(t,\cdot )}{e(t,\cdot )}\Big \Vert _0\le \frac{\eta }{2}\le \frac{r_0}{2}\nonumber \) 

if we take \(\eta =r_0\) , thus
by Lemma  REF
\(e(t,x)Id-R_{0\ell }(t,x)=&e(t,x)\Big (Id-\frac{R_{0\ell }(t,x)}{e(t,x)}\Big )=e(t,x)\sum \limits _{i=1}^{3}\gamma ^2_{k_i}\Big (Id-\frac{R_{0\ell }(t,x)}{e(t,x)}\Big )k_i\otimes k_i\nonumber \\=&\sum \limits _{i=1}^{3}\Big (\rho (t,x)\gamma _{k_i}\Big (Id-\frac{R_{0\ell }(t,x)}{e(t,x)}\Big )\Big )^2k_i\otimes k_i:=\sum \limits _{i=1}^{3}a^2_i(t,x)k_i\otimes k_i.\) 

Where \(a_i(t,x)=\rho (t,x)\gamma _{k_i}\Big (Id-\frac{R_{0\ell }(t,x)}{e(t,x)}\Big )\)  satisfies
\(a_i\in C^{\infty }_c(Q_{r+\delta }),\quad \Vert a_i\Vert _0\le \frac{M\sqrt{\delta }}{300}.\) 

Here we denote constant \(M\)  by
\(M:=\max \Big \lbrace C_0,600\max _{1\le i\le 3}\Vert \gamma _{k_i}\Vert _{L^\infty \big (B_{\frac{r_0}{2}}(Id)\big )}, 600\max _{1\le i\le 3}\frac{1}{\min _{B_{\frac{r_0}{2}}(Id)}\gamma _{k_i}}\Big \rbrace .\) 

Construction of 1-th perturbation on velocity

Main perturbation on velocity
For any \(l\in Z^3\) , we set
\(b_{1l}(t,x):=&\frac{a_1(t,x)\alpha _l(\mu _1t,\mu _1 x)}{\sqrt{2}},\) 

then, by (REF ), it's easy to obtain
\(\Vert b_{1l}\Vert _0\le \frac{M\sqrt{\delta }}{300}.\) 

As in [5]}, we set \([l]:=\sum _{j=0}^22^j[l_j]\) , if \(l=(l_0, l_1, l_2)\) , where
\([l_j]=\left\lbrace \begin{array}{cc}1, ~~~~l_j~~~{\rm is ~~~even},\\0,~~~~ l_j~~~{\rm is~~~ odd}.\end{array}\right.\nonumber \) 

Thus, \([l]\)  can only take values in \(\lbrace 0,1,\cdot \cdot \cdot ,7\rbrace \) .
Now we denote main \(l\) -perturbation \( w_{1ol}\)  by
\(w_{1ol}(t,x):=b_{1l}(t,x)k_1\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^{[l]}k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big ).\) 

Here and subsequent, we denote \(a^{\perp }=(-a_2,a_1)^T\)  if \(a=(a_1,a_2)^T.\)
Then set 1-th main perturbation
\(w_{1o}:=\sum _{l\in Z^3}w_{1ol}=\sum _{j=0}^7\sum _{[l]=j}b_{1l}k_1\Big (e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^jk_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big ).\) 

Obviously, \(w_{1ol},w_{1o}\)  are all real 2-dimensional vector-valued functions.
By (REF ), we have \(\hbox{supp} \alpha _l\cap \hbox{supp}\alpha _{l^{\prime }}=\emptyset \)  if \(|l-l^{\prime }|\ge 2\) , hence there are at most 30 nonzero terms at every point \((t,x)\in R^3\)  in the summation (REF ), thus by (REF )
\(\Vert w_{1o}\Vert _0\le \frac{M\sqrt{\delta }}{6}.\) 

Furthermore, if \(b_{1l}(t,x)\ne 0\) , then \(|(\mu _1t,\mu _1x)-l|\le 1\)  and \(|(t,x)|\le r+\delta \) , thus \(|l|\le C_0(r)\mu _1\) .
By (REF ), we know that for any \(l\in Z^3\) ,
\(b_{1l}\in C^{\infty }_c(Q_{r+\delta }),\quad w_{1ol}\in C^{\infty }_c(Q_{r+\delta }),\quad w_{1o}\in C^{\infty }_c(Q_{r+\delta }).\) 

The correction \(w_{1c}\)  and the 1-th perturbation \(w_{1}\)  
We define \(l\) -correction by
\(w_{1cl}:=&\frac{\nabla ^{\perp }b_{1l}}{i\lambda _1 2^{[l]}}\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}-e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big )\nonumber \\&-\nabla ^{\bot }\Big (\frac{\nabla b_{1l}\cdot k_1^{\perp }}{\lambda _1^22^{2[l]}|k_1|^2}\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big )\Big ),\) 

where \(\nabla ^{\perp }=(-\partial _{x_2},\partial _{x_1})^T.\)
Then 1-th correction is given by
\(w_{1c}:=\sum _{l\in Z^3}w_{1cl}.\) 

Finally, we denote 1-th perturbation \(w_1\)  by
\(w_1:=w_{1o}+w_{1c}.\nonumber \) 

Thus, if we denote \(w_{1l}\)  by
\(w_{1l}:=w_{1ol}+w_{1cl},\nonumber \) 

then, we have
\(w_{1l}=\nabla ^{\perp }{\rm div}\Big (-\frac{b_{1l}}{\lambda _1^22^{2[l]}|k_1|^2}k_1^{\perp }2cos\Big (\lambda _12^{[l]}k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )\Big )\Big )\) 

and
\(w_1=\sum \limits _{l\in Z^3}w_{1l},\quad {\rm div}w_{1l}=0,\quad \int _{R^2}w_{1l}dx=0, \quad \int _{R^2}(x_iw_{1lj}-x_jw_{1li})dx=0, \quad i,j=1,2.\nonumber \) 

In fact
\(\int _{R^2}(x_1w_{1l2}-x_2w_{1l1})dx=\int _{R^2}(x_1\partial _1{\rm div}\vec{a}_1+x_2\partial _2{\rm div}\vec{a}_1)dx=-2\int _{R^2}{\rm div}\vec{a}_1dx=0,\nonumber \) 

where
\(\vec{a}_1=-\frac{b_{1l}}{\lambda _1^22^{2[l]}|k_1|^2}k_1^{\perp }2cos\Big (\lambda _12^{[l]}k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )\Big )\) 
is a vector-valued smooth function with compact support.
Since there are only finite nonzero terms in the summation (REF ) and (REF ), thus
\({\rm div}w_{1}=0,\quad \int _{R^2}w_{1}dx=0, \quad \int _{R^2}(x_iw_{1j}-x_jw_{1i})dx=0, \quad i,j=1,2.\) 

Now we set
\(k_{1l}:=&b_{1l}k_1+\frac{\nabla ^{\perp }b_{1l}}{i\lambda _1 2^{[l]}}-\nabla ^{\bot }\Big (\frac{\nabla b_{1l}\cdot k_1^{\perp }}{\lambda _1^22^{2[l]}|k_1|^2}\Big )+\frac{\nabla b_{1l}\cdot k_1^{\perp }}{i\lambda _12^{[l]}|k_1|^2}\cdot k_1,\nonumber \\k_{-1l}:=&b_{1l}k_1+\frac{\nabla ^{\perp }b_{1l}}{-i\lambda _1 2^{[l]}}-\nabla ^{\bot }\Big (\frac{\nabla b_{1l}\cdot k_1^{\perp }}{\lambda _1^22^{2[l]}|k_1|^2}\Big )+\frac{\nabla b_{1l}\cdot k_1^{\perp }}{-i\lambda _12^{[l]}|k_1|^2}\cdot k_1,\) 

then
\(w_{1l}=k_{1l}e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+k_{-1l}e^{-i\lambda _1 2^{[l]}k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\nonumber \) 

and
\(w_1=\sum _{j=0}^7\sum _{[l]=j}\Big (k_{1l}e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+k_{-1l}e^{-i\lambda _1 2^jk_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big ).\) 

Finally, it's obvious
\(w_{1ol},~w_{1cl},~w_{1l},~w_{1o},~w_{1c},~w_1,~k_{1l},~k_{-1l}\in C^{\infty }_c(Q_{r+\delta }).\nonumber \) 

Thus, we complete the construction of perturbation \(w_1\) .

Construction of 1-th perturbation on temperature
To construct \(\chi _1\) , we first
denote \(\beta _{1l}\)  by
\(\beta _{1l}(t,x):=\frac{c_{1\ell }(t,x)\alpha _1(\mu _1t,\mu _1x)}{\sqrt{2e(t,x)}\gamma _{k_1}\Big (Id-\frac{R_{0\ell }(t,x)}{e(t,x)}\Big )}.\) 

Since \({\rm supp}c_{1\ell }\subseteq {\rm supp}e\) , so \(\beta _{1l}\)  is well-defined.
Then we denote main \(l\) -perturbation \(\chi _{1ol}\)  by
\(\chi _{1ol}(t,x):=&\beta _{1l}(t,x)\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big )\nonumber \) 

and \(l\) -correction \(\chi _{1cl}\)  by
\(\chi _{1cl}(t,x):=&\triangle \beta _{1l}(t,x)\Big (\frac{e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}}{-\lambda _1^22^{2[l]}|k_1|^2}+\frac{e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}}{-\lambda _1^22^{2[l]}|k_1|^2}\Big )\nonumber \\&+2\nabla \beta _{1l}(t,x)\cdot \nabla \Big (\frac{e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}}{-\lambda _1^22^{2[l]}|k_1|^2}+\frac{e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}}{-\lambda _1^22^{2[l]}|k_1|^2}\Big ).\nonumber \) 

Finally, the \(l\) -th perturbation is given by
\(\chi _{1l}(t,x):=\chi _{1ol}(t,x)+\chi _{1cl}(t,x)=\triangle \Big (\beta _{1l}(x,t)\Big (\frac{e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}}{-\lambda _1^22^{2[l]}|k_1|^2}+\frac{e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}}{-\lambda _1^22^{2[l]}|k_1|^2}\Big )\Big ).\nonumber \) 

Set
\(\chi _{1o}(t,x):=\sum _{l\in Z^3}\chi _{1ol}(t,x),\quad \chi _{1c}(t,x):=\sum _{l\in Z^3}\chi _{1cl}(t,x),\quad \chi _1(t,x):=\sum _{l\in Z^3}\chi _{1l}(t,x).\nonumber \) 

Obviously, \(\chi _{1ol}, ~\chi _{1cl}, ~\chi _{1l}\)  and \(\chi _1\)  are all real scalar functions and
\(\chi _{1o}(t,x)=\sum _{j=0}^7\sum _{[l]=j}\beta _{1l}(t,x)\Big (e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big ).\nonumber \) 

Moreover, it's easy to get
\(\int _{R^2} \chi _{1l}(t,x)dx=0,\qquad \int _{R^2} x_1\chi _{1l}(t,x)dx=0.\nonumber \) 

Since there are only finite terms in the summation of \(\chi _1\) , therefore
\(\int _{R^2} \chi _{1}(t,x)dx=0,\qquad \int _{R^2} x_1\chi _{1}(t,x)dx=0.\) 

If set
\(h_{1l}:=\beta _{1l}-\frac{\triangle \beta _{1l}}{\lambda _1^22^{2[l]}|k_1|^2}+2\frac{\nabla \beta _{1l}\cdot k_1^{\perp }}{i\lambda _1 2^{[l]}|k_1|^2},\quad h_{-1l}:=\beta _{1l}-\frac{\triangle \beta _{1l}}{\lambda _1^22^{2[l]}|k_1|^2}+2\frac{\nabla \beta _{1l}\cdot k_1^{\perp }}{-i\lambda _1 2^{[l]}|k_1|^2},\) 

then
\(\chi _{1l}=h_{1l}e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+h_{-1l}e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\nonumber \) 

and
\(\chi _{1}=\sum _{j=0}^7\sum _{[l]=j}\Big (h_{1l}e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+h_{-1l}e^{-i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big ).\) 

Since \(c_{1\ell }(t,x)\in C^{\infty }_c(Q_{r+\delta })\) , we know that for all \(l\in Z^3\)
\(\beta _{1l}\in C^{\infty }_c(Q_{r+\delta }),\qquad h_{1l}\in C^{\infty }_c(Q_{r+\delta })\nonumber \) 

and
\(\chi _{1ol},\quad \chi _{1cl},\quad \chi _{1l},\quad \chi _1\in C^{\infty }_c(Q_{r+\delta }).\nonumber \) 

Then, by (REF ), (REF ), (REF ), (REF ) and (REF ), we know that
\(\Vert \beta _{1l}\Vert _0\le \frac{M\sqrt{\delta }}{200}.\nonumber \) 

Similar to (REF ), we have
\(\Vert \chi _{1o}\Vert _0\le \frac{M\sqrt{\delta }}{4}.\) 

The construction of \(v_{01}\) , \(p_{01}\) , \(\theta _{01}\) , \(R_{01}\) , \(f_{01}\)  
First, we denote
\(M_1\)  by
\(M_1:=\sum \limits _{l\in Z^3}b^2_{1l}k_1\otimes k_1\Big (e^{2i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+e^{-2i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big {)}+\sum \limits _{l,l^{\prime }\in Z^3 ,l\ne l^{\prime }}w_{1ol}\otimes w_{1ol^{\prime }}\nonumber \) 
and
\(N_1,K_1\)  by
\(N_1:=&\sum _{l\in Z^3}\Big [w_{1l}\otimes \Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )+\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big {)}\otimes w_{1l}\Big ]+R_0-R_{0\ell },\nonumber \\K_1:=&\sum \limits _{l\in Z^3}\beta _{1l}b_{1l}k_1\Big (e^{2i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+e^{-2i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big {)}+\sum \limits _{l,l^{\prime }\in Z^3 ,l\ne l^{\prime }}w_{1ol}\chi _{1ol^{\prime }}.\nonumber \) 

Then set
\(v_{01}(t,x):=v_0(t,x)+w_1(t,x),\quad p_{01}(t,x):=&p_0(t,x)-e(t,x),\quad \theta _{01}(t,x):=\theta _0(t,x)+\chi _1(t,x),\nonumber \\R_{01}(t,x):=-\bar{R}_{0\ell }(t,x)+&2\sum \limits _{l\in Z^3}b^2_{1l}(t,x)k_1\otimes k_1+\delta R_{01}(t,x),\nonumber \\f_{01}(t,x):=f_{0\ell }(t,x)+2&\sum \limits _{l\in Z^3}\beta _{1l}(t,x)b_{1l}(t,x)k_1+\delta f_{01}(t,x),\) 

where
\(\bar{R}_{0\ell }(t,x)=-R_{0\ell }(t,x)+e(t,x)Id,\nonumber \) 
\(\delta R_{01}:=&\mathcal {R}(\hbox{div}M_1)+N_1-\mathcal {R}(\chi _1e_2)+\mathcal {R}\Big \lbrace \partial _tw_{1}+\hbox{div}\Big [\sum _{l\in Z^3}\Big (w_{1l}\otimes v_0\Big (\frac{l}{\mu _{1}}\Big )+v_0\Big (\frac{l}{\mu _{1}}\Big )\otimes w_{1l}\Big )\Big ]\Big \rbrace \nonumber \\&+(w_{1o}\otimes w_{1c}+w_{1c}\otimes w_{1o}+w_{1c}\otimes w_{1c})\) 

and
\(\delta f_{01}:=&\mathcal {G}(\hbox{div}K_1)+\mathcal {G}\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \chi _{1l}\Big )+w_{1o}\chi _{1c}+f_0-f_{0\ell }+\sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}\nonumber \\&+w_{1c}\chi _1+\sum _{l\in Z^3}w_{1l}\Big (\theta _0-\theta _0\Big (\frac{l}{\mu _2}\Big )\Big ).\) 

By (REF ) and (REF ), we know
\(\hbox{div}M_1,~\chi _1e_2, ~\partial _tw_{1},~\hbox{div}\Big [\sum _{l\in Z^3}\Big (w_{1l}\otimes v_0\Big (\frac{l}{\mu _{1}}\Big )+v_0\Big (\frac{l}{\mu _{1}}\Big )\otimes w_{1l}\Big )\Big ]\in \Xi ,\) 
so \(\delta R_{01}\)  is well-defined. Notice that
\(\hbox{div}K_1,~\partial _t\chi _{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \chi _{1l}\in \Psi ,\) 
thus \(\delta f_{01}\)  is also well-defined.
By Proposition REF , we know that \(\delta R_{01}\)  is a symmetric matrix and \(\delta R_{01}\in C_c^{\infty }(Q_{r+\delta })\) . Also, by Proposition REF , we have \(\delta f_{01}\in C_c^{\infty }(Q_{r+\delta })\) .
Obviously,
\(\hbox{div}v_{01}=\hbox{div}v_0+\hbox{div}w_{1}=0.\) 

Moreover, from the definition of \((v_{01}, p_{01}, \theta _{01}, R_{01}, f_{01})\)  and the fact that \((v_{0}, p_{0},\theta _0, R_{0}, f_{0})\)  solves the system (REF ), together with Proposition REF  we know that
\(\begin{aligned}\hbox{div}R_{01}=&\hbox{div}R_0-\nabla e+\partial _tw_1-\chi _1e_2+\hbox{div}(w_{1o}\otimes w_{1o}+w_{1}\otimes v_{0}+v_0\otimes w_{1}\nonumber \\&+w_{1o}\otimes w_{1c}+w_{1c}\otimes w_{1o}+w_{1c}\otimes w_{1c})\nonumber \\=&\partial _tv_{0}+\hbox{div}(v_{0}\otimes v_{0})+\nabla p_{0}-\theta _0e_2-\nabla e+\partial _tw_1-\chi _1e_2+\hbox{div}(w_{1o}\otimes w_{1o}\nonumber \\&+w_{1}\otimes v_{0}+v_0\otimes w_{1}+w_{1o}\otimes w_{1c}+w_{1c}\otimes w_{1o}+w_{1c}\otimes w_{1c})\nonumber \\=&\partial _tv_{01}+\hbox{div}(v_{01}\otimes v_{01})+\nabla p_{01}-\theta _{01}e_2.\end{aligned}\) 
Here we use the fact
\(\hbox{div}(M_1)+\hbox{div}\Big (2\sum \limits _{l\in Z^3}b^2_{1l}(x,t)k_1\otimes k_1)\Big )=\hbox{div}(w_{1o}\otimes w_{1o}).\) 
Furthermore, by (REF ) and (REF ), we have
\( \begin{aligned}f_{01}=&f_0+2\sum \limits _{l\in Z^3}\beta _{1l}b_{1l}k_1+\mathcal {G}(\hbox{div}K_1)+w_{1o}\chi _{1c}+\mathcal {G}\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \chi _{1l}\Big )\nonumber \\&+\sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _1}\Big )\Big )\chi _{1l}+w_{1c}\chi _1+\sum _{l\in Z^3}w_{1l}\Big (\theta _0-\theta _0\Big (\frac{l}{\mu _2}\Big )\Big ).\nonumber \end{aligned}\) 
Thus, by Proposition REF  and the fact that \((v_{0}, p_{0},\theta _0, R_{0}, f_{0})\)  solves the system (REF ), we have
\(\hbox{div}f_{01}=&\hbox{div}f_0+\partial _{t}\chi _1+\hbox{div}(w_{1o}\chi _1+w_{1c} \chi _1+v_0 \chi _1+w_1 \theta _0)\nonumber \\=&\hbox{div}(v_0\theta _0+w_{1o}\chi _1+w_{1c} \chi _1+v_0 \chi _1+w_1 \theta _0)+\partial _t(\theta _0+\chi _1)\nonumber \\=&\partial _t\theta _{01}+\hbox{div}(v_{01}\theta _{01}).\nonumber \) 

Here we use the fact
\(\hbox{div}K_1+\hbox{div}(2\sum \limits _{l\in Z^3}\beta _{1l}(x,t)b_{1l}(x,t)k_1)=\hbox{div}(w_{1o}\chi _{1o}).\nonumber \) 

Thus the new functions \((v_{01},p_{01},\theta _{01},R_{01},f_{01})\)  solves the system (REF ).

The 1-th representations
In this section, we will calculate the following two terms
\(-\bar{R}_{0\ell }+2\sum \limits _{l\in Z^3}b^2_{1l}k_1\otimes k_1=I\) 
and
\(f_{0\ell }+2\sum \limits _{l\in Z^3}\beta _{1l}b_{1l}k_1=II.\) 
The term I
First, by (REF ) and the fact \(\sum \limits _{l\in Z^3}\alpha _l^2=1\) , we have
\(2\sum \limits _{l\in Z^3}b^2_{1l}(t,x)k_1\otimes k_1=\sum \limits _{l\in Z^3}\alpha _l^2(\mu _1t,\mu _1 x) a_1^2(t,x)k_1\otimes k_1=a_1^2(t,x)k_1\otimes k_1.\nonumber \) 

Thus, by (REF ),
\(-\bar{R}_{0\ell }(x,t)+2\sum \limits _{l\in Z^3}b^2_{1l}(x,t)k_1\otimes k_1=-\sum _{i=2}^3 a_i^2(t,x)k_i\otimes k_i.\nonumber \) 

Meanwhile, we have
\(R_{01}=-\sum _{i=2}^3 a_i^2(t,x)k_i\otimes k_i+\delta R_{01}.\) 

In next section, we will prove that \(\delta R_{01}\)  is small.

The term II
By (REF ) and (REF ), we have
\(2\sum \limits _{l\in Z^3}\beta _{1l}(t,x)b_{1l}(t,x)k_1=\sum \limits _{l\in Z^3}\alpha _l^2(\mu _1t,\mu _1 x)c_1(t,x)k_1=c_1(t,x)k_1.\nonumber \) 

By (REF ),
\(f_{0\ell }+2\sum \limits _{l\in Z^3}\beta _{1l}(x,t)b_{1l}(x,t)k_1=-c_2(t,x)k_2.\nonumber \) 

Meanwhile, we have
\(f_{01}(t,x)=-c_2(t,x)k_2+\delta f_{01}.\) 

Again in next section, we will prove that \(\delta f_{01}\)  is small.

Estimates on \(\delta R_{01}\)  and \(\delta f_{01}\)
In the subsequent estimates, unless otherwise stated, \(C_0\)  always denotes an absolute constant which only depends on \(r, \Vert v_0\Vert _0\)  and \(C_m\)  will in addition to depend on \(m\) 
and both them can vary from line to line.
In the following, we frequently use the elementary inequality
\([fg]_{m}\le & C_m\bigl ([f]_{m}\Vert g\Vert _0+[g]_{m}\Vert f\Vert _0\bigr )\) 

for any \(m\ge 0\) .
Moreover, by the standard estimates on convolution,
\(\Vert c_{i\ell }\Vert _{C^m_{t,x}}+\Vert R_{0\ell }\Vert _{C^m_{t,x}}\le & C_m\Lambda \ell ^{1-m}\quad {\rm for~~any~~m\ge 1},~~~ i=1,2,3.\\\Vert R_{0\ell }-R_0\Vert _0+\Vert f_{0\ell }-f_0\Vert _0\le & C_0\Lambda \ell .\) 

Then, we collect a classical estimates on the Hölder norms of compositions, their proof can be found in [6]}:
Let \(u: R^n\rightarrow R^N\)  and \(\Psi : R^N\rightarrow R\)  be two smooth functions. Then, for every \(m\in {\rm N}\setminus {0}\)  there is a constant \(C_m=C_m(N,n)\)  such that
\([\Psi (u)]_m\le & C_m\sum \limits _{i=1}^m[\Psi ]_i[u]_1^{(i-1)\frac{m}{m-1}}[u]_m^{\frac{m-i}{m-1}}.\) 

In particular,
\([\Psi (u)]_1\le C_1[\Psi ]_1[u]_1.\) 
We summarize the main estimates on \(b_{1l}\)  and \(\beta _{1l}\)  in the following lemma.
Lemma 6.1 
For any \(l\in Z^3\)  and integer \(m\ge 1\) , we have
\(\Vert b_{1l}\Vert _m+\Vert \beta _{1l}\Vert _m\le & C_m\sqrt{\delta }(\mu _1^m+\mu _1\ell ^{-(m-1)}),\\\Vert \partial _tb_{1l}\Vert _m+\Vert \partial _t\beta _{1l}\Vert _m\le & C_m\sqrt{\delta }(\mu _1^{m+1}+\mu _1\ell ^{-m}),\\\Vert \partial _{tt}b_{1l}\Vert _m+\Vert \partial _{tt}\beta _{1l}\Vert _m\le & C_m\sqrt{\delta }(\mu _1^{m+2}+\mu _1\ell ^{-m-1}),\\\Vert k_{\pm 1l}\Vert _m+\Vert h_{\pm 1l}\Vert _m\le & C_m\sqrt{\delta }(\mu _1^m+\mu _1\ell ^{-(m-1)}),\\\Vert \partial _tk_{\pm 1l}\Vert _m+\Vert \partial _th_{\pm 1l}\Vert _m\le & C_m\sqrt{\delta }(\mu _1^{m+1}+\mu _1\ell ^{-m}),\\\Vert \partial _{tt}k_{\pm 1l}\Vert _m+\Vert \partial _{tt}h_{\pm 1l}\Vert _m\le &C_m\sqrt{\delta }(\mu _1^{m+2}+\mu _1\ell ^{-m-1})\) 

and
\(\Vert b_{1l}\Vert _0+\Vert \beta _{1l}\Vert _0\le & C_0\sqrt{\delta },\\\Vert \partial _tb_{1l}\Vert _0+\Vert \partial _t\beta _{1l}\Vert _0\le & C_0\sqrt{\delta }\mu _1,\\\Vert \partial _{tt}b_{1l}\Vert _0+\Vert \partial _{tt}\beta _{1l}\Vert _0\le & C_0\sqrt{\delta }(\mu _1^{2}+\mu _1\ell ^{-1}),\\\Vert k_{\pm 1l}\Vert _0+\Vert h_{\pm 1l}\Vert _0\le & C_0\sqrt{\delta },\\\Vert \partial _tk_{\pm 1l}\Vert _0+\Vert \partial _th_{\pm 1l}\Vert _0\le & C_0\sqrt{\delta }\mu _1,\\\Vert \partial _{tt}k_{\pm 1l}\Vert _0+\Vert \partial _{tt}h_{\pm 1l}\Vert _0\le &C_0\sqrt{\delta }(\mu _1^{2}+\mu _1\ell ^{-1}).\) 
First, notice the fact \(\lbrace (t,x)|\nabla e\ne 0\rbrace \cap \lbrace (t,x)|R_{0\ell }\ne 0\rbrace =\emptyset \) , we have, for any positive integer \(i\) ,
\(\nabla ^i\Big (\frac{R_{0\ell }}{e}\Big )=\frac{\nabla ^i( R_{0\ell })}{e},\) 
thus for \(m\ge 1\) , by (REF ), (REF ), (REF ), (REF ) and parameter assumption (REF )
\(\Big [\gamma _{k_1}\Big (Id-\frac{R_{0\ell }}{e}\Big )(t,\cdot )\Big ]_m\le & C_m\sum \limits _{i=1}^m\Vert \nabla ^i\gamma _{k_1}\Vert _0\Big [Id-\frac{R_{0\ell }}{e}(t,\cdot )\Big ]_1^{(i-1)\frac{m}{m-1}}\Big [Id-\frac{R_{0\ell }}{e}(t,\cdot )\Big ]_m^{\frac{m-i}{m-1}}\nonumber \\\le & C_m(\mu _1^m+\mu _1\ell ^{-(m-1)}).\nonumber \) 

It's obvious that
\(\Big \Vert \gamma _{k_1}\Big (Id-\frac{R_{0\ell }}{e}\Big )\Big \Vert _0\le & C_0.\nonumber \) 

Moreover, by (REF ) and parameter assumption (REF ), for any integer \(m\) ,
\(\Vert \rho \Vert _{C^m_{t,x}}\le C_m\delta ^{-\frac{1}{2}-(m-1)}\le C_m\sqrt{\delta }\mu _1^m.\nonumber \) 

Then, recalling that
\(b_{1l}(t,x):=\frac{\rho (t,x)}{\sqrt{2}}\gamma _{k_1}\Big (Id-\frac{R_{0\ell }}{e}\Big )(t,x)\alpha _l(\mu _1t,\mu _1x),\nonumber \) 

thus, for \(m\ge 1\) , by (REF ) and parameter assumption (REF ),
it's easy to get
\(\sup _t\big [b_{1l}(t,\cdot )\big ]_m\le C_m\sqrt{\delta }(\mu _1^m+\mu _1\ell ^{-(m-1)}).\nonumber \) 

By (REF ) and parameter assumption (REF ), for any integer \(m\ge 1\) ,
\(\Vert c_{1\ell }\Vert _{C^m_{t,x}}\le C_m\Lambda \ell ^{-(m-1)}\le C_m\delta \mu _1\ell ^{-(m-1)}.\) 
By (REF ), \(\Vert c_{1\ell }\Vert _0\le 2\delta \) . By (REF ), (REF ), (REF ) and parameter assumption (REF ), for any integer \(m\) ,
\(\Big \Vert \frac{1}{\sqrt{e}}\Big \Vert _{C^m_{t,x}}\le C_m\delta ^{-\frac{1}{2}-m}\le C_m\delta ^{-\frac{1}{2}}\mu _1^m.\nonumber \) 

Notice that \(\frac{1}{2}\le \gamma _{k_1}\le \frac{3}{2}\) , then, by (REF ), (REF ) and parameter assumption (REF ), we also have
\(\Vert \beta _{1l}\Vert _m\le C_m\sqrt{\delta }(\mu _1^m+\mu _1\ell ^{-(m-1)}).\nonumber \) 

Thus, we complete the proof of (REF ). (REF ) is a direct result of (REF ) and ().
By the definition (REF ) on \(k_{\pm 1l}\) , the definition (REF ) on \(h_{\pm 1l}\) , estimate (REF ) and parameter assumption (REF ), for \(m\ge 1\) , it's easy to obtain
\(\Vert k_{\pm 1l}\Vert _m\le C_m\sqrt{\delta }(\mu _1^m+\mu _1\ell ^{-(m-1)}),\quad \Vert h_{\pm 1l}\Vert _m\le C_m\sqrt{\delta }(\mu _1^m+\mu _1\ell ^{-(m-1)}).\nonumber \) 

Thus, we complete the proof of (). The case of \(m=0\)  in () is also direct.
We introduce function
\(\Gamma (t,x)=\frac{\rho (t,x)}{\sqrt{2}}\gamma _{k_1}\Big (Id-\frac{R_{0\ell }}{e}\Big )(t,x).\) 
By (REF ), (REF ) and parameter assumption (REF ) and notice the fact
\(\partial _t\Big (\frac{R_{0\ell }}{e}\Big )=\frac{(\partial _t R_0)_{\ell }}{e}-\frac{R_{0\ell }\partial _te}{e^2}=\frac{(\partial _t R_0)_{\ell }}{e},\quad \partial _{tt}\Big (\frac{R_{0\ell }}{e}\Big )=\frac{\partial _t R_0\ast (\partial _t\varphi )_{\ell }\ell ^{-1}}{e}\) 
we have
\(\Vert \partial _t\Gamma \Vert _0\le C_0\sqrt{\delta }\mu _1,\quad \Vert \partial _{tt}\Gamma \Vert _0\le C_0\sqrt{\delta }(\mu _1^2+\mu _1\ell ^{-1}).\nonumber \) 

Moreover, by (REF ), (REF ), (REF ), (REF ) and parameter assumption (REF ), for \(m\ge 1\)  we have
\(\Vert \partial _t\Gamma \Vert _m\le C_m\sqrt{\delta }(\mu _1^{m+1}+\mu _1\ell ^{-m}),\quad \Vert \partial _{tt}\Gamma \Vert _m\le C_m\sqrt{\delta }(\mu _1^{m+2}+\mu _1\ell ^{-m-1}).\nonumber \) 

Observe that
\(b_{1l}(t,x)=\Gamma (t,x)\alpha _l(\mu _1t, \mu _1x),\) 
thus
\(\partial _tb_{1l}=&\partial _t\Gamma \alpha _l(\mu _1t, \mu _1x)+\mu _1\Gamma (\partial _t\alpha )_l(\mu _1t,\mu _1x),\nonumber \\\partial _{tt}b_{1l}=&\partial _{tt}\Gamma \alpha _l(\mu _1t, \mu _1x)+2\mu _1\partial _t\Gamma (\partial _t\alpha )_l(\mu _1t, \mu _1x)+\mu _1^2\Gamma (\partial _{tt}\alpha )_l(\mu _1t,\mu _1x).\nonumber \) 

Hence, by (REF ) and the above estimate on \(\Gamma \) , we obtain
\(\Vert \partial _tb_{1l}\Vert _0\le C_0\sqrt{\delta }\mu _1,\quad \Vert \partial _{tt}b_{1l}\Vert _0\le C_0\sqrt{\delta }(\mu _1^2+\mu _1\ell ^{-1})\) 

and
\(\Vert \partial _tb_{1l}\Vert _m\le C_m\sqrt{\delta }(\mu _1\ell ^{-m}+\mu _1^{m+1}),\quad \Vert \partial _{tt}b_{1l}\Vert _m\le C_m\sqrt{\delta }(\mu _1\ell ^{-m-1}+\mu _1^{m+2}).\nonumber \) 

The same argument gives
\(\Vert \partial _t\beta _{1l}\Vert _0\le & C_0\sqrt{\delta }\mu _1,\quad \quad \quad \Vert \partial _{tt}\beta _{1l}\Vert _0\le C_0\sqrt{\delta }(\mu _1^2+\mu _1\ell ^{-1}),\\\Vert \partial _t\beta _{1l}\Vert _m\le & C_m\sqrt{\delta }(\mu _1\ell ^{-m}+\mu _1^{m+1}),\quad \Vert \partial _{tt}\beta _{1l}\Vert _m\le C_m\sqrt{\delta }(\mu _1\ell ^{-m-1}+\mu _1^{m+2}).\nonumber \) 

Thus, we obtain (), (), () and ().
Then, by the definition (REF ) on \(k_{\pm 1l}\) , the definition (REF ) on \(h_{\pm 1l}\)  and parameter assumption (REF ), it's easy to obtain (), (), () and ().
Thus, the proof of this lemma is complete.
MM\(\square \)
Next, we give estimates on perturbations \(w_{1o}, w_{1c}, \chi _{1o}, \chi _{1c}\) .
Lemma 6.2 (Estimate on main perturbation and correction)
\(\Vert w_{1o}\Vert _0\le &C_0\sqrt{\delta },\quad \Vert w_{1o}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\lambda _1,\quad \Vert \chi _{1o}\Vert _0\le C_0\sqrt{\delta },\quad \Vert \chi _{1o}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\lambda _1,\\\Vert w_{1c}\Vert _0\le & C_0\frac{\sqrt{\delta }\mu _1}{\lambda _1},\quad \Vert w_{1c}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\mu _1,\quad \Vert \chi _{1c}\Vert _0\le C_0\frac{\sqrt{\delta }\mu _1}{\lambda _1},\quad \Vert \chi _{1c}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\mu _1.\) 
First, by (REF ) and (REF ), we know \(\Vert w_{1o}\Vert _0\le C_0\sqrt{\delta }, \quad \Vert \chi _{1o}\Vert _0\le C_0\sqrt{\delta }\) . Since
\(\partial _tw_{1o}=&\sum _{l\in Z^3}\partial _tb_{1l}k_1\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big ((y,z)-v_0(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big ((y,z)-v_0(\frac{l}{\mu _1})t\big )}\Big )\nonumber \\&-\sum _{l\in Z^3}b_{1l}i\lambda _1 2^{[l]} k_1^{\perp }\cdot v_0\Big (\frac{l}{\mu _1}\Big )k_1\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big ((y,z)-v_0(\frac{l}{\mu _1})t\big )}-e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big ((y,z)-v_0(\frac{l}{\mu _1})t\big )}\Big ),\nonumber \) 

thus, by (REF ), () and parameter assumption (REF ), we obtain
\(\Vert \partial _tw_{1o}\Vert _0\le C_0\sqrt{\delta }\lambda _1.\nonumber \) 

The same argument gives
\(\Vert \nabla w_{1o}\Vert _0\le C_0\sqrt{\delta }\lambda _1,\quad \Vert \chi _{1o}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\lambda _1.\nonumber \) 

Thus, we give a proof of (REF ).
Next, by (REF ), (REF ), parameter assumption (REF ), we get
\(\Vert w_{1cl}\Vert _0\le C_0\frac{\sqrt{\delta }\mu _1}{\lambda _1}.\nonumber \) 

Thus, from the property (REF ) on \(\alpha _l\) , we arrive at
\(\Vert w_{1c}\Vert _0\le C_0\frac{\sqrt{\delta }\mu _1}{\lambda _1}.\nonumber \) 

Differentiating (REF ) in time
\(\partial _tw_{1cl}=&\sum _{l\in Z^3}\Big \lbrace \frac{\nabla ^{\perp }\partial _tb_{1l}}{i\lambda _1 2^{[l]}}\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}-e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big )\nonumber \\&-\nabla ^{\perp }b_{1l}k_1^{\perp }\cdot v_{0}\Big (\frac{l}{\mu _1}\Big )\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big )\nonumber \\&-\nabla ^{\bot }\Big (\frac{\nabla \partial _tb_{1l}\cdot k_1^{\perp }}{\lambda _1^22^{2[l]}|k_1|^2}\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big )\nonumber \\&-\nabla ^{\bot }\Big (\frac{\nabla b_{1l}\cdot k_1^{\perp }}{i\lambda _12^{[l]}|k_1|^2}k_1^{\perp }\cdot v_{0}\Big (\frac{l}{\mu _1}\Big )\Big (e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}-e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big )\Big \rbrace .\nonumber \) 

By (REF ), () and parameter assumption (REF ), we get
\(\Vert \partial _tw_{1c}\Vert _0\le C_0\sqrt{\delta }\mu _1.\nonumber \) 

Similarly, we have
\(\Vert \nabla w_{1c}\Vert _0\le C_0\sqrt{\delta }\mu _1,\quad \Vert \chi _{1c}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\mu _1.\nonumber \) 

Collect the above estimates, we complete the proof of ().
MM\(\square \)
By (REF ), (REF ), (REF ), (REF ), (REF ) and lemma REF , it's easy to obtain the following estimate:
Corollary 6.3
\(\Vert v_{01}-v_0\Vert _0\le \frac{M\sqrt{\delta }}{6}+C_0\frac{\sqrt{\delta }\mu _1}{\lambda _1},\quad \Vert p_{01}-p_0\Vert _0\le & M\delta ,\quad \Vert \theta _{01}-\theta _0\Vert _0\le \frac{M\sqrt{\delta }}{4}+C_0\frac{\sqrt{\delta }\mu _1}{\lambda _1},\nonumber \\\Vert v_{01}-v_0\Vert _{C^1_{t,x}}\le C_0\lambda _1\sqrt{\delta },\qquad \Vert p_{01}-p_0\Vert _{C^1_{t,x}}\le & C_0,\quad \Vert \theta _{01}-\theta _0\Vert _{C^1_{t,x}}\le C_0\lambda _1\sqrt{\delta }.\) 

Estimates on \(\delta R_{01}\)
Recalling that
\(\delta R_{01}=&\mathcal {R}(\hbox{div}M_1)+N_1-\mathcal {R}(\chi _1e_2)+\mathcal {R}\Big \lbrace \partial _tw_{1}+\hbox{div}\Big [\sum _{l\in Z^3}\Big (w_{1l}\otimes v_0\Big (\frac{l}{\mu _{1}}\Big )+v_0\Big (\frac{l}{\mu _{1}}\Big )\otimes w_{1l}\Big )\Big ]\Big \rbrace \nonumber \\&+(w_{1o}\otimes w_{1c}+w_{1c}\otimes w_{1o}+w_{1c}\otimes w_{1c}).\nonumber \) 

We split the stress into three parts:
(1)The oscillation part
\(\mathcal {R}(\hbox{div}M_1)-\mathcal {R}(\chi _1e_2).\) 
(2)The transport part
\(\mathcal {R}\Big \lbrace \partial _tw_{1}+\hbox{div}\Big [\sum _{l\in Z^3}\Big (w_{1l}\otimes v_0\Big (\frac{l}{\mu _{1}}\Big )+v_0\Big (\frac{l}{\mu _{1}}\Big )\otimes w_{1l}\Big )\Big ]\Big \rbrace =\mathcal {R}\Big (\partial _tw_{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla w_{1l}\Big ).\nonumber \) 

(3)The error part
\(N_1+(w_{1o}\otimes w_{1c}+w_{1c}\otimes w_{1o}+w_{1c}\otimes w_{1c}).\nonumber \) 

In the following we will estimate each term separately.
Lemma 6.4 (The oscillation part)
\(\Vert \mathcal {R}({\rm div}M_1)\Vert _0\le & C_0(\varepsilon )\frac{\delta \mu _1}{\lambda _1},\quad \Vert \mathcal {R}({\rm div}M_1)\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\delta \mu _1.\\\Vert \mathcal {R}(\chi _1e_2)\Vert _0\le & C_0(\varepsilon )\frac{\sqrt{\delta }}{\lambda _1},\quad \Vert \mathcal {R}(\chi _1e_2)\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sqrt{\delta }.\) 
First, we have
\(M_{1}=\sum _{j=0}^7\sum _{[l]=j}k_1\otimes k_1\Big (e^{2i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+e^{-2i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big {)}b^2_{1l}+\sum \limits _{l,l^{\prime }\in Z^3 ,l\ne l^{\prime }}w_{1ol}\otimes w_{1ol^{\prime }}.\nonumber \) 

Since \(k_1\cdot k_1^{\perp }=0,\)  then
\(\hbox{div}M_1=M_{11}+M_{12}.\nonumber \) 

where
\(M_{11}=&\sum _{j=0}^7\sum _{[l]=j}k_1\otimes k_1\nabla (b^2_{1l})\Big (e^{2i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+e^{-2i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big ),\nonumber \\M_{12}=&\sum \limits _{l,l^{\prime }\in Z^3 ,l\ne l^{\prime }}\hbox{div}(w_{1ol}\otimes w_{1ol^{\prime }}).\nonumber \) 

By (REF ), (REF ), Proposition REF  with \(m=\Big [1+\frac{1}{\varepsilon }\Big ]+1\)  and parameter assumption (REF ) , we have
\(\Vert \mathcal {R}(M_{11})\Vert _0\le & C_m\sum \limits _{j=0}^7\Big (\sum _{i=0}^{m-1}\frac{\Vert \sum _{[l]=j}\nabla (b^2_{1l})\Vert _i}{(\lambda _12^j)^{i+1}}+\frac{\Vert \sum _{[l]=j}\nabla (b^2_{1l})\Vert _m}{(\lambda _12^j)^{m}}\Big )\nonumber \\\le & C_m\sum \limits _{j=0}^7\delta \Big (\frac{\mu _1}{\lambda _12^j}+\frac{\mu _1^{m+1}+\mu _1\ell ^{-m}}{(\lambda _12^j)^m}\Big )\le C_m\frac{\delta \mu _1}{\lambda _1}.\nonumber \) 

where we use the fact: \(b_{1l}b_{1l^{\prime }}= 0\)  if \(|l-l^{\prime }|\ge 2\) .
Notice
\(M_{12}=&\sum \limits _{j=0}^7\sum _{[l]=j}\sum \limits _{l^{\prime }\in Z^3, 1\le |l^{\prime }-l|<2}k_1\otimes k_1\nabla (b_{1l}b_{1l^{\prime }})\Big (e^{i\lambda _1(2^j+2^{[l^{\prime }]})k_1^{\perp }\cdot x-ig_{1,l,l^{\prime }}(t)}+e^{i\lambda _1(2^j-2^{[l^{\prime }]})k_1^{\perp }\cdot x-i\overline{g}_{1,l,l^{\prime }}(t)}\nonumber \\&+e^{i\lambda _1(2^{[l^{\prime }]}-2^j)k_1^{\perp }\cdot x+i\overline{g}_{1,l,l^{\prime }}(t)}+e^{-i\lambda _1(2^j+2^{[l^{\prime }]})k_1^{\perp }\cdot x+ig_{1,l,l^{\prime }}(t)}\Big ),\nonumber \) 

where
\(g_{1,l,l^{\prime }}(t)=\lambda _1\Big (2^{[l]}k_1^{\perp }\cdot v_0\Big (\frac{l}{\mu _1}\Big )t+2^{[l^{\prime }]}k_1^{\perp }\cdot v_0\Big (\frac{l^{\prime }}{\mu _1}\Big )t\Big ),~~~~~\overline{g}_{1,l,l^{\prime }}(t)=\lambda _1\Big (2^{[l]}k_1^{\perp }\cdot v_0\Big (\frac{l}{\mu _1}\Big )-2^{[l^{\prime }]}k_1^{\perp }\cdot v_0\Big (\frac{l^{\prime }}{\mu _1}\Big )\Big ).\) 
Following the same strategy as \(M_{11}\) , we deduce
\(\Vert \mathcal {R}(M_{12})\Vert _0\le & C_m\sum \limits _{j=0}^7\Big (\sum _{i=0}^{m-1}\frac{\Big \Vert \sum _{[l]=j}\sum \limits _{l^{\prime }\in Z^3, 1\le |l^{\prime }-l|<2} \nabla (b_{1l}b_{1l^{\prime }})\Big \Vert _i}{\lambda _1^{i+1}}+\frac{\Big \Vert \sum _{[l]=j}\sum \limits _{l^{\prime }\in Z^3, 1\le |l^{\prime }-l|<2}\nabla (b_{1l}b_{1l^{\prime }})\Big \Vert _m}{\lambda _1^{m}}\Big )\nonumber \\\le & C_m\sum \limits _{j=0}^7\delta \Big (\frac{\mu _1}{\lambda _1}+\frac{\mu _1^{m+1}+\mu _1\ell ^{-m}}{\lambda _1^m}\Big )\le C_m\frac{\delta \mu _1}{\lambda _1}.\) 

Thus, the first estimate of (REF ) follows easily.
A direct calculation gives
\(\partial _tM_{11}=&\sum _{j=0}^7\sum _{[l]=j}k_1\otimes k_1\nabla \partial _t(b^2_{1l})\Big (e^{2i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+e^{-2i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big )\nonumber \\&-\sum _{j=0}^7\sum _{[l]=j}k_1\otimes k_1\nabla (b^2_{1l})2i\lambda _1 2^j k_1^{\perp }\cdot v_0\Big (\frac{l}{\mu _1}\Big )\Big (e^{2i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}-e^{-2i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big ).\nonumber \) 

Thus, by (REF ), (), (REF ), Proposition REF  with \(m=\Big [1+\frac{1}{\varepsilon }\Big ]+1\)  and parameter assumption (REF ), we have
\(\Vert \partial _t\mathcal {R}(M_{11})\Vert _0\le & C_m\sum \limits _{j=0}^7\Big (\sum _{i=0}^{m-1}\frac{\Vert \sum _{[l]=j}\nabla \partial _t(b^2_{1l})\Vert _i}{(\lambda _12^j)^{i+1}}+\frac{\Vert \sum _{[l]=j}\nabla \partial _t(b^2_{1l})\Vert _m}{(\lambda _12^j)^{m}}\Big )\nonumber \\&+ C_m\lambda _1\sum \limits _{j=0}^7\Big (\sum _{i=0}^{m-1}\frac{\Vert \sum _{[l]=j}\nabla (b^2_{1l})\Vert _i}{(\lambda _12^j)^{i+1}}+\frac{\Vert \sum _{[l]=j}\nabla (b^2_{1l})\Vert _m}{(\lambda _12^j)^{m}}\Big )\nonumber \\\le & C_m\lambda _1\sum \limits _{j=0}^7\delta \Big (\frac{\mu _1}{\lambda _12^j}+\frac{\mu _1^{m+1}+\mu _1\ell ^{-m}}{(\lambda _12^j)^m}\Big )\le C_m\delta \mu _1.\nonumber \) 

By the same argument, we have
\(\Vert \partial _t\mathcal {R}(M_{12})\Vert _0\le C_m\delta \mu _1,\quad \Vert \nabla \mathcal {R}(M_{11})\Vert _0\le C_m\delta \mu _1,\quad \Vert \nabla \mathcal {R}(M_{12})\Vert _0\le C_m\delta \mu _1.\nonumber \) 

Putting these estimates together, we obtain the second estimate of (REF ).
Recalling (REF )
\(\chi _{1}=\sum _{j=0}^7\sum _{[l]=j}\Big (h_{1l}e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}+h_{-1l}e^{-i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _1})t\big )}\Big ).\nonumber \) 

By (), (), (), () and using a similar argument as above, we obtain
\(\big \Vert \mathcal {R}(\chi _1e_2)\big \Vert _0\le C_m\frac{\sqrt{\delta }}{\lambda _1},\quad \big \Vert \mathcal {R}(\chi _1e_2)\big \Vert _{C^1_{t,x}}\le C_m\sqrt{\delta }.\nonumber \) 

Thus, the proof of this lemma is complete.
MM\(\square \)
Lemma 6.5 (The transport part)
\(\Big \Vert \mathcal {R}\Big (\partial _tw_{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla w_{1l}\Big )\Big \Vert _0\le C_0(\varepsilon )\frac{\sqrt{\delta }\mu _1}{\lambda _1},\nonumber \\\Big \Vert \mathcal {R}\Big (\partial _tw_{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla w_{1l}\Big )\Big \Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sqrt{\delta }\mu _1.\) 
Recalling (REF )
\(w_1=\sum _{j=0}^7\sum _{[l]=j}\Big (k_{1l}e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+k_{-1l}e^{-i\lambda _1 2^jk_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big ).\nonumber \) 

Thus, using the identity
\(\Big (\partial _t+v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \Big )e^{\pm i\lambda _1 2^{[l]} k_1^{\perp }\cdot (x-v_{0}(\frac{l}{\mu _1})t\big )}=0,\) 
we deduce
\(\partial _tw_{1}+\sum _{l\in Z^3}v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla w_{1l}=&\sum _{j=0}^7\sum _{[l]=j}\Big (\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )k_{1l}e^{i\lambda _1 2^j k_1^{\perp }\cdot (x-v_{0}(\frac{l}{\mu _1})t\big )}\nonumber \\&+\Big (\partial _t+v_0\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )k_{-1l}e^{-i\lambda _1 2^j k_1^{\perp }\cdot (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big {)}.\nonumber \) 

By Proposition REF  with \(m=\Big [1+\frac{1}{\varepsilon }\Big ]+1\)  , (), (), () and parameter assumption(REF ), we have
\(&\Big \Vert \mathcal {R}\Big (\partial _tw_{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla w_{1l}\Big )\Big \Vert _0\nonumber \\\le & C_m\sum \limits _{j=0}^7\Big (\sum _{i=0}^{m-1}\frac{\Big \Vert \sum _{[l]=j}\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )k_{1l}\Big \Vert _i}{(\lambda _12^j)^{i+1}}+\frac{\Big \Vert \sum _{[l]=j}\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )k_{1l}\Big \Vert _m}{(\lambda _12^j)^{m}}\Big )\nonumber \\&+C_m\sum \limits _{j=0}^7\Big (\sum _{i=0}^{m-1}\frac{\Big \Vert \sum _{[l]=j}\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )k_{-1l}\Big \Vert _i}{(\lambda _12^j)^{i+1}}+\frac{\Big \Vert \sum _{[l]=j}\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )k_{-1l}\Big \Vert _m}{(\lambda _12^j)^{m}}\Big )\nonumber \\\le & C_m\sum \limits _{j=0}^7\sqrt{\delta }\Big (\frac{\mu _1}{\lambda _12^j}+\frac{\mu _1^{m+1}+\mu _1\ell ^{-m}}{(\lambda _12^j)^m}\Big )\le C_m\frac{\sqrt{\delta }\mu _1}{\lambda _1}.\) 

A direct calculation gives
\(\partial _t\Big (\partial _tw_{1}+\sum _{l\in Z^3}v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla w_{1l}\Big )\nonumber =&\sum _{j=0}^7\sum _{[l]=j}\Big (\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )\partial _tk_{1l}e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\nonumber \\&+\Big (\partial _t+v_0\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )\partial _tk_{-1l}e^{-i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\nonumber \\&-\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )k_{1l}i\lambda _1 2^j k_1^{\perp }\cdot v_{0}\Big (\frac{l}{\mu _1}\Big )e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\nonumber \\&+\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )k_{-1l}i\lambda _1 2^j k_1^{\perp }\cdot v_{0}\Big (\frac{l}{\mu _1}\Big )e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big ),\nonumber \) 

thus, by ()-(), ()-() and applying the same argument as above, we arrive at
\(\Big \Vert \partial _t\mathcal {R}\Big (\partial _tw_{1}+\sum _{l\in Z^3}v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla w_{1l}\Big )\Big \Vert _0\le C_m\sqrt{\delta }\mu _1.\nonumber \) 

Similarly, we have
\(\Big \Vert \nabla \mathcal {R}\Big (\partial _tw_{1}+\sum _{l\in Z^3}v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla w_{1l}\Big )\Big \Vert _0\le C_m\sqrt{\delta }\mu _1.\nonumber \) 

Then we proved the Lemma REF .
MM\(\square \)
Lemma 6.6 (Estimates on error part I)
\(\Vert N_1\Vert _0\le C_0\Big (\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ),\qquad \Vert N_1\Vert _{C^1_{t,x}}\le C_0\lambda _1\Big (\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ).\) 

We may rewrite \(N_{1}\)  as
\(N_1=N_{11}+N_{12},\nonumber \) 

where
\(N_{11}=\sum _{l\in Z^3}\Big [w_{1l}\otimes \Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )+\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big {)}\otimes w_{1l}\Big ],\quad N_{12}=R_0-R_{0\ell }.\nonumber \) 

For the term \(N_{11}\) , by (REF ), we have
\(&\sum _{l\in Z^3}w_{1l}\otimes \Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\nonumber \\=&\sum _{l\in Z^3}\Big (k_{1l}\otimes \Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+k_{-1l}\otimes \Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )e^{-i\lambda _1 2^{[l]}k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big ).\nonumber \) 

Obviously, \(k_{1l}(x,t)\ne 0\)  implies \(|(\mu _1t,\mu _1 x)-l|\le 1\) , therefore, by ()
\(\Big |k_{1l}\otimes \Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\Big |\le C_0\sqrt{\delta }\frac{\Vert \nabla v_0\Vert _0}{\mu _1}\le C_0\sqrt{\delta }\frac{\Lambda }{\mu _1}.\) 
By (REF ), it's easy to get
\(\Big \Vert \sum _{l\in Z^3}k_{1l}\otimes \Big (v_0-v_{0}\Big (\frac{l}{\mu _1}\Big )\Big )\Big \Vert _0\le C_0\sqrt{\delta }\frac{\Lambda }{\mu _1}.\nonumber \) 

Similarly,
\(\Big \Vert \sum _{l\in Z^3} k_{-1l}\otimes \Big (v_0-v_{0}\Big (\frac{l}{\mu _1}\Big )\Big )\Big \Vert _0\le C_0\sqrt{\delta }\frac{\Lambda }{\mu _1}.\nonumber \) 

Thus,
\(\Big \Vert \sum _{l\in Z^3} w_{1l}\otimes \Big (v_0-v_{0}\Big (\frac{l}{\mu _1}\Big )\Big )\Big \Vert _0\le C_0\sqrt{\delta }\frac{\Lambda }{\mu _1}.\) 

Following the same strategy:
\(\Big \Vert \sum _{l\in Z^3}\Big (v_0-v_{0}\Big (\frac{l}{\mu _1}\Big )\Big )\otimes w_{1l}\Big \Vert _0\le C_0\sqrt{\delta }\frac{\Lambda }{\mu _1}.\) 

Finally, putting (REF ) and (REF ) together, we arrive at
\(\Vert N_{11}\Vert _0\le C_0\sqrt{\delta }\frac{\Lambda }{\mu _1}.\) 

Moreover, we have
\(&\partial _t\Big (\sum _{l\in Z^3}w_{1l}\otimes \Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\Big )\nonumber \\=&\sum _{l\in Z^3}\Big (\partial _tk_{1l}e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+\partial _tk_{-1l}e^{-i\lambda _1 2^{[l]}k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big {)}\otimes \Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\nonumber \\&+\sum _{l\in Z^3}\Big (k_{1l}e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+k_{-1l}e^{-i\lambda _1 2^{[l]}k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big {)}\otimes \partial _tv_0\nonumber \\&+\sum _{l\in Z^3}\Big (-k_{1l}i\lambda _1 2^{[l]} k_1^{\perp }\cdot v_{0}\Big (\frac{l}{\mu _1}\Big )e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\nonumber \\&+k_{-1l}i\lambda _1 2^{[l]} k_1^{\perp }\cdot v_{0}\Big (\frac{l}{\mu _1}\Big )e^{-i\lambda _1 2^{[l]}k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big {)}\otimes \Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big ),\nonumber \) 

thus, by (), () and parameter assumption(REF )
\(\Big \Vert \partial _t\Big (\sum _{l\in Z^3}w_{1l}\otimes \Big (v_0-v_{0}\Big (\frac{l}{\mu _1}\Big )\Big )\Big )\Big \Vert _0\le C_0\sqrt{\delta }\lambda _1\frac{\Lambda }{\mu _1}.\nonumber \) 

Similarly, we have
\(\Big \Vert \partial _t\Big (\sum _{l\in Z^3}\Big (v_0-v_{0}\Big (\frac{l}{\mu _1}\Big )\Big )\otimes w_{1l}\Big )\Big \Vert _0\le C_0\sqrt{\delta }\lambda _1\frac{\Lambda }{\mu _1}.\nonumber \) 

Therefore,
\(\Vert \partial _tN_{11}\Vert _0\le C_0\sqrt{\delta }\lambda _1\frac{\Lambda }{\mu _1}.\) 

By a similar argument, we have
\(\Vert \nabla N_{11}\Vert _0\le C_0\sqrt{\delta }\lambda _1\frac{\Lambda }{\mu _1}.\) 

By (), we have
\(\Vert R_0-R_{0\ell }\Vert _0\le C_0\Lambda \ell ,\quad \Vert \partial _t(R_0-R_{0\ell })\Vert _0\le C_0\Lambda ,\quad \Vert \nabla (R_0-R_{0,\ell })\Vert _0\le C_0\Lambda .\nonumber \) 

Thus, by parameter assumption(REF ), we arrive at
\(\Vert N_{12}\Vert _0\le C_0\Lambda \ell ,\quad \Vert N_{12}\Vert _{C^1_{t,x}}\le C_0\lambda _1\Lambda \ell .\) 

Collecting (REF ), (REF ), (REF ) and
(REF ), we obtain
\(\Vert N_1\Vert _0\le C_0\Big (\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ),\qquad \Vert N_1\Vert _{C^1_{t,x}}\le C_0\lambda _1\Big (\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ).\nonumber \) 

We complete our proof of this lemma.
MM\(\square \)
Lemma 6.7 (Estimates on error part II)
\(\Vert w_{1o}\otimes w_{1c}+w_{1c}\otimes w_{1o}+w_{1c}\otimes w_{1c}\Vert _0\le C_0\frac{\delta \mu _1}{\lambda _1},\nonumber \\\Vert w_{1o}\otimes w_{1c}+w_{1c}\otimes w_{1o}+w_{1c}\otimes w_{1c}\Vert _{C^1_{t,x}}\le C_0\delta \mu _1.\) 
By (REF ) and (), we have
\(\Vert w_{1o}\otimes w_{1c}+w_{1c}\otimes w_{1o}+w_{1c}\otimes w_{1c}\Vert _0\le C_0(\Vert w_{1o}\Vert _0\Vert w_{1c}\Vert _0+\Vert w_{1c}\Vert _0^2)\le C_0\frac{\delta \mu _1}{\lambda _1}\nonumber \) 

and
\(\Vert w_{1o}\otimes w_{1c}+w_{1c}\otimes w_{1o}+w_{1c}\otimes w_{1c}\Vert _{C^1_{t,x}}\le C_0(\Vert w_{1o}\Vert _0\Vert w_{1c}\Vert _{C^1_{t,x}}+\Vert w_{1o}\Vert _{C^1_{t,x}}\Vert w_{1c}\Vert _0)\le C_0\delta \mu _1.\nonumber \) 

thus, we complete the proof of this lemma.
MM\(\square \)
Finally, from Lemma REF , Lemma REF , Lemma REF  and Lemma REF , we conclude
\(\Vert \delta R_{01}\Vert _0\le C_0(\varepsilon )\Big (\frac{\sqrt{\delta }\mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ),\quad \Vert \delta R_{01}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _1\Big (\frac{\sqrt{\delta }\mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ).\) 
Estimates on \(\delta f_{01}\)
Recalling that
\(\delta f_{01}=&\mathcal {G}(\hbox{div}K_1)+\mathcal {G}\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \chi _{1l}\Big )+w_{1o}\chi _{1c}+f_0-f_{0\ell }\nonumber \\&+\sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}+w_{1c}\chi _1+\sum _{l\in Z^3}w_{1l}\Big (\theta _0-\theta _0\Big (\frac{l}{\mu _1}\Big )\Big ).\nonumber \) 

As before, we split \(\delta f_{01}\)  into three parts:
(1)The oscillation part:
\(\mathcal {G}(\hbox{div}K_1).\nonumber \) 

(2)The transport part:
\(\mathcal {G}\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \chi _{1l}\Big ).\nonumber \) 

(3)The error part:
\(w_{1c}\chi _1+w_{1o}\chi _{1c}+f_0-f_{0\ell }+\sum \limits _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}+\sum _{l\in Z^3}w_{1l}\Big (\theta _0-\theta _0\Big (\frac{l}{\mu _2}\Big )\Big ).\nonumber \) 
Lemma 6.8 (The oscillation part)
\(\Vert \mathcal {G}({\rm div}K_1)\Vert _0\le C_0(\varepsilon )\frac{\delta \mu _1}{\lambda _1},\qquad \Vert \mathcal {G}({\rm div}K_1)\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\delta \mu _1.\) 
Recalling the notations of \(K_1\)  and \([l]\) , we have
\(K_1=\sum _{j=0}^7\sum _{[l]=j}\beta _{1l}b_{1l}k_1\Big (e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^jk_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big {)}\Big )+\sum \limits _{l,l^{\prime }\in Z^3 ,l\ne l^{\prime }}w_{1ol}\chi _{1ol^{\prime }},\nonumber \) 

thus
\(\hbox{div}K_1=K_{11}+K_{12},\nonumber \) 

where
\(K_{11}=&\sum _{j=0}^7\sum _{[l]=j}\nabla (\beta _{1l}b_{1l})\cdot k_1\Big (e^{i\lambda _1 2^{|l|} k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^{|l|}k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big {)},\nonumber \\K_{12}=&\sum \limits _{l,l^{\prime }\in Z^3 ,l\ne l^{\prime }}\hbox{div}(w_{1ol}\chi _{1ol^{\prime }}).\nonumber \) 

By (REF ), (REF ), Proposition (REF ) with \(m=\Big [1+\frac{1}{\varepsilon }\Big ]+1\)  and parameter assumption (REF ) , we have
\(\Vert \mathcal {R}(K_{11})\Vert _0\le & C_m\sum \limits _{j=0}^7\Big (\sum _{i=0}^{m-1}\frac{\Vert \sum _{[l]=j}\nabla (\beta _{1l}b_{1l})\Vert _i}{(\lambda _12^j)^{i+1}}+\frac{\Vert \sum _{[l]=j}\nabla (\beta _{1l}b_{1l})\Vert _m}{(\lambda _12^j)^{m}}\Big )\nonumber \\\le & C_m\sum \limits _{j=0}^7\delta \Big (\frac{\mu _1}{\lambda _12^j}+\frac{\mu _1^{m+1}+\mu _1\ell ^{-m}}{(\lambda _12^j)^m}\Big )\le C_m\frac{\delta \mu _1}{\lambda _1}.\nonumber \) 

Moreover, we have
\(K_{12}=&\sum \limits _{j=0}^7\sum _{[l]=j}\sum \limits _{l^{\prime }\in Z^3,1\le |l^{\prime }-l|<2}k_1\cdot \nabla (b_{1l}\beta _{1l^{\prime }})\Big (e^{i\lambda _1(2^j+2^{[l^{\prime }]})k_1^{\perp }\cdot x-ig_{1,l,l^{\prime }}(t)}+e^{i\lambda _1(2^j-2^{[l^{\prime }]})k_1^{\perp }\cdot x-i\overline{g}_{1,l,l^{\prime }}(t)}\nonumber \\&+e^{i\lambda _1(2^{[l^{\prime }]}-2^{j})k_1^{\perp }\cdot x+i\overline{g}_{1,l,l^{\prime }}(t)}+e^{-i\lambda _1(2^j+2^{[l^{\prime }]})k_1^{\perp }\cdot x+ig_{1,l,l^{\prime }}(t)}\Big ),\nonumber \) 

as estimate (REF ) on \(M_{12}\) , by (REF ), (REF ) and Proposition (REF ), we have
\(\Vert \mathcal {G}(K_{12})\Vert _0\le C_m\frac{\delta \mu _1}{\lambda _1}.\nonumber \) 

A straightforward computation gives
\(\partial _tK_{11}=&\sum _{j=0}^7\sum _{[l]=j}\nabla \partial _t(\beta _{1l}b_{1l})\cdot k_1\Big (e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+e^{-i\lambda _1 2^jk_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big )\nonumber \\&-\sum _{j=0}^7\sum _{[l]=j}\nabla (\beta _{1l}b_{1l})\cdot k_{1}i\lambda _1 2^j k_1^{\perp }\cdot v_{0}\Big (\frac{l}{\mu _1}\Big )\Big (e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}-e^{-i\lambda _1 2^jk_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big {)},\nonumber \) 

thus, by the same argument
\(\Vert \partial _t\mathcal {G}(K_{11})\Vert _0\le C_m\delta \mu _1.\nonumber \) 

Similarly,
\(\Vert \partial _t\mathcal {G}(K_{12})\Vert _0\le C_m\delta \mu _1,\quad \Vert \nabla \mathcal {G}(K_{11})\Vert _0\le C_m\delta \mu _1,\qquad \Vert \nabla \mathcal {G}( K_{12})\Vert _0\le C_m\delta \mu _1.\nonumber \) 

We complete the proof of this lemma.
MM\(\square \)
Lemma 6.9 (The transport part)
\(\Big \Vert \mathcal {G}\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \chi _{1l}\Big )\Big \Vert _0\le C_0(\varepsilon ) \frac{\sqrt{\delta }\mu _1}{\lambda _1},\nonumber \\\Big \Vert \mathcal {G}\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \chi _{1l}\Big )\Big \Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sqrt{\delta } \mu _1.\) 
Recalling the notation of \(\chi _1\) , we have
\(&\partial _t\chi _{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \chi _{1l}\nonumber \\=&\sum _{j=0}^7\sum _{[l]=j}\Big \lbrace \Big (\partial _t+v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \Big )h_{1l}e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}+\Big (\partial _t+v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \Big )h_{-1l}e^{-i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big \rbrace .\nonumber \) 

By Proposition REF  with \(m=\Big [1+\frac{1}{\varepsilon }\Big ]+1\)  , (), (), () and parameter assumption(REF ), as estimate in (REF ) , we have
\(\Big \Vert \mathcal {G}\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_0\Big (\frac{l}{\mu _{1}}\Big )\cdot \nabla \chi _{1l}\Big )\Big \Vert _0\le C_m\sum \limits _{j=0}^7\sqrt{\delta }\Big (\frac{\mu _1}{\lambda _12^j}+\frac{\mu _1^{m+1}+\mu _1\ell ^{-m}}{(\lambda _12^j)^m}\Big )\le C_m\frac{\sqrt{\delta }\mu _1}{\lambda _1}.\nonumber \) 

And since
\(\partial _t\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \chi _{1l}\Big )\nonumber =&\sum _{j=0}^7\sum _{[l]=j}\Big (\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )\partial _th_{1l}e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\nonumber \\&+\Big (\partial _t+v_0\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )\partial _th_{-1l}e^{-i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\nonumber \\&-\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )h_{1l}i\lambda _1 2^j k_1^{\perp }\cdot v_{0}\Big (\frac{l}{\mu _1}\Big )e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\nonumber \\&+\Big (\partial _t+v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \Big )h_{-1l}i\lambda _1 2^j k_1^{\perp }\cdot v_{0}\Big (\frac{l}{\mu _1}\Big )e^{i\lambda _1 2^j k_1^{\perp }\cdot \big (x-v_{0}(\frac{l}{\mu _1})t\big )}\Big ),\nonumber \) 

then, by the same argument, we obtain
\(\Big \Vert \partial _t\mathcal {G}\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \chi _{1l}\Big )\Big \Vert _0\le C_m\sqrt{\delta }\mu _1.\nonumber \) 

Similarly, we have
\(\Big \Vert \nabla \mathcal {G}\Big (\partial _t\chi _{1}+\sum _{l\in Z^3}v_{0}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla \chi _{1l}\Big )\Big \Vert _0\le C_m\sqrt{\delta }\mu _1.\nonumber \) 

Then we complete our proof of this Lemma.
MM\(\square \)
Lemma 6.10 (The error part)
\(&\Big \Vert w_{1c}\chi _1+w_{1o}\chi _{1c}+f_0-f_{0\ell }+\sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}+\sum _{l\in Z^3}w_{1l}\Big (\theta _0-\theta _0\Big (\frac{l}{\mu _1}\Big )\Big )\Big \Vert _0\nonumber \\\le & C_0\Big (\frac{\delta \mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ),\nonumber \\&\Big \Vert w_{1c}\chi _1+w_{1o}\chi _{1c}+f_0-f_{0\ell }+\sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}+\sum _{l\in Z^3}w_{1l}\Big (\theta _0-\theta _0\Big (\frac{l}{\mu _1}\Big )\Big )\Big \Vert _{C^1_{t,x}}\nonumber \\\le & C_0\lambda _1\Big (\frac{\delta \mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ).\) 
Using Lemma REF , it's easy to obtain
\(\Vert w_{1c}\chi _1\Vert _0\le C_0\frac{\delta \mu _1}{\lambda _1},\quad \Vert w_{1o}\chi _{1c}\Vert _0\le C_0\frac{\delta \mu _1}{\lambda _1},\quad \Vert w_{1c}\chi _1\Vert _{C^1_{t,x}}\le C_0\delta \mu _1,\quad \Vert w_{1o}\chi _{1c}\Vert _{C^1_{t,x}}\le C_0\delta \mu _1.\nonumber \) 

Then,
\(&\sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}\nonumber \\=&\sum \limits _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )h_{1l}e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _{1}})t\big )}+\sum \limits _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )h_{-1l}e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _{1}})t\big )}.\nonumber \) 

Obviously, \(h_{1l}(x,t), h_{-1l}\ne 0\)  implies \(|(\mu _1t,\mu _1 x)-l|\le 1.\) 
Therefore, following the same strategy as estimate (REF ), we obtain
\(\Big \Vert \sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}\Big \Vert _0\le C_0\sqrt{\delta }\frac{\Lambda }{\mu _1}.\nonumber \) 

Similarly, we have
\(\Big \Vert \sum _{l\in Z^3}w_{1l}\Big (\theta _0-\theta _0\Big (\frac{l}{\mu _1}\Big )\Big )\Big \Vert _0\le C_0\sqrt{\delta }\frac{\Lambda }{\mu _1}.\nonumber \) 

By calculation we have
\(&\partial _t\Big (\sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}\Big )\nonumber \\=&\sum \limits _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\partial _th_{1l}e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _{1}})t\big )}+\sum \limits _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\partial _th_{-1l}e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _{1}})t\big )}\nonumber \\&-\sum \limits _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )h_{1l}i\lambda _1 2^{[l]} k_1^{\perp }\cdot v_0\Big (\frac{l}{\mu _{1}}\Big )e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _{1}})t\big )}\nonumber \\&+\sum \limits _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )h_{-1l}i\lambda _1 2^{[l]} k_1^{\perp }\cdot v_0\Big (\frac{l}{\mu _{1}}\Big )e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _{1}})t\big )}\nonumber \\&+\sum \limits _{l\in Z^3}\partial _tv_0h_{1l}e^{i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _{1}})t\big )}+\sum \limits _{l\in Z^3}\partial _tv_0h_{-1l}e^{-i\lambda _1 2^{[l]} k_1^{\perp }\cdot \big (x-v_0(\frac{l}{\mu _{1}})t\big )}.\nonumber \) 

Therefore, by (), () and parameter assumption(REF )
\(\Big \Vert \partial _t\Big (\sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}\Big )\Big \Vert _0\le C_0\sqrt{\delta }\lambda _1\frac{\Lambda }{\mu _1}.\nonumber \) 

Similarly, we have
\(\Big \Vert \nabla \Big (\sum _{l\in Z^3}\Big (v_0-v_0\Big (\frac{l}{\mu _{1}}\Big )\Big )\chi _{1l}\Big )\Big \Vert _0\le C_0\sqrt{\delta }\lambda _1\frac{\Lambda }{\mu _1}.\nonumber \) 

Applying the similar argument, we have
\(\Big \Vert \sum _{l\in Z^3}w_{1l}\Big (\theta _0-\theta _0\Big (\frac{l}{\mu _1}\Big )\Big )\Big \Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\lambda _1\frac{\Lambda }{\mu _1}.\nonumber \) 

By ()
\(\Vert f_0-f_{0\ell }\Vert _0 \le C_0\Lambda \ell ,\qquad \Vert f_0-f_{0\ell }\Vert _{C^1_{t,x}} \le C_0\Lambda .\nonumber \) 

Collecting the above estimates together, we complete our proof.
MM\(\square \)
Combining lemma REF , lemma REF  and lemma REF , we conclude
\(\Vert \delta f_{01}\Vert _0\le C_0(\varepsilon )\Big (\frac{\sqrt{\delta }\mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ),\quad \Vert \delta f_{01}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _1\Big (\frac{\sqrt{\delta }\mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ).\) 

Finally, by (REF ), (REF ), Corollary REF , (REF ) and (REF ), we conclude that \((v_{01},p_{01},\theta _{01}, R_{01}, f_{01})\in C_c^{\infty }(Q_{r+\delta })\)  solves system (REF ) and satisfies
\(R_{01}(t,x)=-\sum _{i=2}^3 a^2_i(t,x)k_i\otimes k_i+\delta R_{01},\qquad f_{01}(t,x):=-c_2(t,x)k_2+\delta f_{01}\nonumber \) 

with
\(\Vert v_{01}-v_0\Vert _0\le \frac{M\sqrt{\delta }}{6}+C_0\frac{\sqrt{\delta }\mu _1}{\lambda _1},\quad \Vert p_{01}-p_0\Vert _0\le & M\delta ,\quad \Vert \theta _{01}-\theta _0\Vert _0\le \frac{M\sqrt{\delta }}{4}+C_0\frac{\sqrt{\delta }\mu _1}{\lambda _1},\nonumber \\\Vert v_{01}-v_0\Vert _{C^1_{t,x}}\le C_0\lambda _1\sqrt{\delta },\qquad \Vert p_{01}-p_0\Vert _{C^1_{t,x}}\le & C_0,\qquad \Vert \theta _{01}-\theta _0\Vert _{C^1_{t,x}}\le C_0\lambda _1\sqrt{\delta },\nonumber \\\Vert \delta R_{01}\Vert _0\le C_0(\varepsilon )\Big (\sqrt{\delta }\frac{\mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ),\quad &\Vert \delta R_{01}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _1\Big (\sqrt{\delta }\frac{\mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ),\nonumber \\\Vert \delta f_{01}\Vert _0\le C_0(\varepsilon )\Big (\sqrt{\delta }\frac{\mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ),\quad &\Vert \delta f_{01}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _1\Big (\sqrt{\delta }\frac{\mu _1}{\lambda _1}+\sqrt{\delta }\frac{\Lambda }{\mu _1}+\Lambda \ell \Big ).\nonumber \) 

Thus, we complete the first step.
Constructions of \((v_{0n}, p_{0n}, \theta _{0n}, R_{0n}, f_{0n})\) , \(2\le n\le 3\)
In this section, we suppose \(2\le n \le 3\)  and construct \((v_{0n}, p_{0n}, \theta _{0n}, R_{0n}, f_{0n})\)  by inductions.
Suppose that for \(1\le m<n\le 3\) , \((v_{0m}, p_{0m}, \theta _{0m}, R_{0m}, f_{0m})\in C_c^{\infty }(Q_{r+\delta })\)  solves system (REF ) and satisfies
\(R_{0m}=-\sum _{i=m+1}^3 a^2_ik_i\otimes k_i+\sum _{i=i}^m\delta R_{0i},\quad f_{0m}:=-\sum _{i=m+1}^3c_ik_i+\sum _{i=i}^m\delta f_{0i}\) 

with
\(\Vert v_{0m}-v_{0(m-1)}\Vert _0\le \frac{M\sqrt{\delta }}{6}+C_0\frac{\sqrt{\delta }\mu _m}{\lambda _m}& ,\quad \Vert v_{0m}-v_{0(m-1)}\Vert _{C^1_{t,x}}\le C_0\lambda _m\sqrt{\delta },\nonumber \\\Vert p_{0m}-p_{0(m-1)}\Vert _0\le \left\lbrace \begin{array}{ll}M\delta , m=1\\0, \quad m=2\end{array}\right.&\quad \Vert p_{0m}-p_{0(m-1)}\Vert _{C^1_{t,x}}\le \left\lbrace \begin{array}{cc}C_0, m=1\\0, m=2\end{array}\right.\nonumber \\\Vert \theta _{0m}-\theta _{0(m-1)}\Vert _0\le \frac{M\sqrt{\delta }}{6}+C_0\frac{\sqrt{\delta }\mu _m}{\lambda _m},&\quad \Vert \theta _{0m}-\theta _{0(m-1)}\Vert _{C^1_{t,x}}\le C_0\lambda _m\sqrt{\delta } .\) 

and
\(\Vert \delta R_{0m}\Vert _0\le C_0(\varepsilon )\Big (\frac{\sqrt{\delta }\mu _m}{\lambda _m}+\sqrt{\delta }\frac{\sqrt{\delta }\lambda _{m-1}}{\mu _m}\Big ),\quad &\Vert \delta f_{0m}\Vert _0\le C_0(\varepsilon )\Big (\frac{\sqrt{\delta }\mu _m}{\lambda _m}+\sqrt{\delta }\frac{\sqrt{\delta }\lambda _{m-1}}{\mu _m}\Big ),\nonumber \\\Vert \delta R_{0m}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _m\Big (\frac{\sqrt{\delta }\mu _m}{\lambda _m}+\sqrt{\delta }\frac{\sqrt{\delta }\lambda _{m-1}}{\mu _m}\Big ),\quad &\Vert \delta f_{0m}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _m\Big (\frac{\sqrt{\delta }\mu _m}{\lambda _m}+\sqrt{\delta }\frac{\sqrt{\delta }\lambda _{m-1}}{\mu _m}\Big ).\) 

Here \((v_{00}, p_{00}, \theta _{00})=(v_0, p_0, \theta _0)\)  and the parameter \(\mu _m, \lambda _m\)  satisfies
\(\lambda _m\ge \max \lbrace \mu _m^{1+\varepsilon },\ell ^{-(1+\varepsilon )}\rbrace , \quad \mu _m>\mu _{m-1}\) 

and \(\lambda _0=\Lambda \delta ^{-\frac{1}{2}}+\frac{\mu _1\Lambda \ell }{\delta }, \mu _0=1\) .
Next, we perform the n-th step.
Construction of n-th perturbation on velocity

Main perturbation \(w_{no}\)
For any \(l\in Z^3\) , we denote \(b_{nl}\)  by
\(b_{nl}(t,x):=&\frac{a_n(t,x)\alpha _l(\mu _nt,\mu _nx)}{\sqrt{2}}\) 

and main \(l\) -perturbation \(w_{1ol}\)  by
\(w_{nol}:=b_{nl}k_n\Big (e^{i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}+e^{-i\lambda _n 2^{[l]}k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}\Big ),\) 

where two parameters \(\mu _n\)  and \(\lambda _n\)  will be chosen with
\(\lambda _n\ge \max \lbrace \mu _n^{1+\varepsilon },\ell ^{-(1+\varepsilon )}\rbrace , \quad \mu _n>\mu _{n-1}.\) 

Then we denote n-th main perturbation \(w_{no}\)  by
\(w_{no}:=\sum _{l\in Z^3}w_{nol}=\sum _{j=0}^7\sum _{[l]=j}b_{nl}k_n\Big (e^{i\lambda _n 2^j k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}+e^{-i\lambda _n 2^jk_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}\Big ).\nonumber \) 

Obviously, \(w_{nol},w_{no}\)  are all real 2-dimensional vector-valued functions.
By (REF ), \(\hbox{supp} \alpha _l\cap \hbox{supp}\alpha _{l^{\prime }}=\emptyset \)  if \(|l-l^{\prime }|\ge 2\) , thus the above summation is finite and
\(\Vert w_{no}\Vert _0\le \frac{M\sqrt{\delta }}{6}.\) 

Moreover, since \(a_n(t,x)\in C^{\infty }_c(Q_{r+\delta })\) , we know that for any \(l\in Z^3\) ,
\(b_{nl}\in C^{\infty }_c(Q_{r+\delta }),\quad w_{nol}\in C^{\infty }_c(Q_{r+\delta })\quad w_{no}\in C^{\infty }_c(Q_{r+\delta }).\) 

The correction \(w_{nc}\)  and the n-th perturbation \(w_{n}\)
We denote the \(l\) -correction \(w_{ncl}\)  by
\(w_{ncl}:=&\frac{\nabla ^{\perp }b_{nl}}{i\lambda _n 2^{[l]}}\Big (e^{i\lambda _n 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}-e^{-i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}\Big )\nonumber \\&-\nabla ^{\bot }\Big (\frac{\nabla b_{nl}\cdot k_n^{\perp }}{\lambda _n^22^{2[l]}|k_n|^2}\Big (e^{i\lambda _n 2^{[l]} k_1^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}+e^{-i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}\Big )\Big ).\nonumber \) 

Then the n-th correction is given by
\(w_{nc}:=\sum _{l\in Z^3}w_{ncl}.\nonumber \) 

Finally, we denote n-th perturbation \(w_n\)  by
\(w_n:=w_{no}+w_{nc}.\nonumber \) 

Thus, if we set
\(w_{nl}:=w_{nol}+w_{ncl},\nonumber \) 

then
\(w_{nl}=\nabla ^{\perp }{\rm div}\Big (-\frac{b_{nl}}{\lambda _n^22^{2[l]}|k_n|^2}k_n^{\perp }2cos\Big (\lambda _n2^{[l]}k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )\Big )\Big )\nonumber \) 

and
\(w_n=\sum \limits _{l\in Z^3}w_{nl},\quad {\rm div}w_{nl}=0,\quad \int _{R^2}w_{nl}dx=0, \quad \int _{R^2}(x_iw_{nlj}-x_jw_{nli})dx=0,\quad i.j=1,2.\nonumber \) 

In fact
\(\int _{R^2}(x_1w_{nl2}-x_2w_{nl1})dx=\int _{R^2}(x_1\partial _1{\rm div}\vec{a}_n+x_2\partial _2{\rm div}\vec{a}_n)dx=-2\int _{R^2}{\rm div}\vec{a}_ndx=0,\nonumber \) 

where
\(\vec{a}_n=-\frac{b_{nl}}{\lambda _n^22^{2[l]}|k_n|^2}k_1^{\perp }2cos\Big (\lambda _n2^{[l]}k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )\Big )\) 
is a vector-valued smooth function with compact support.
Obviously, we have
\(\hbox{div}w_n=0.\nonumber \) 

Moreover, if we set
\(k_{nl}:=&b_{nl}k_n+\frac{\nabla ^{\perp }b_{nl}}{i\lambda _n 2^{[l]}}-\nabla ^{\bot }\Big (\frac{\nabla b_{nl}\cdot k_n^{\perp }}{\lambda _n^22^{2[l]}|k_n|^2}\Big )+\frac{\nabla b_{nl}\cdot k_n^{\perp }}{i\lambda _n2^{[l]}|k_n|^2}\cdot k_n,\nonumber \\k_{-nl}:=&b_{nl}k_n+\frac{\nabla ^{\perp }b_{nl}}{-i\lambda _n 2^{[l]}}-\nabla ^{\bot }\Big (\frac{\nabla b_{nl}\cdot k_n^{\perp }}{\lambda _n^22^{2[l]}|k_n|^2}\Big )+\frac{\nabla b_{nl}\cdot k_n^{\perp }}{-i\lambda _n2^{[l]}|k_n|^2}\cdot k_n,\nonumber \) 

then
\(w_{nl}=k_{nl}e^{i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}+k_{-nl}e^{-i\lambda _n 2^{[l]}k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}.\) 
Furthermore, we have
\(w_{nol},~w_{ncl},~w_{nl},~w_{no},~w_{nc},~w_n,~k_{nl},~k_{-nl}\in C^{\infty }_c(Q_{r+\delta }).\nonumber \) 

Thus, we complete the construction of perturbation \(w_n\) .

Construction of n-th perturbation on temperature
To construct \(\chi _n\) , we denote \(\beta _{nl}\)  by
\(\beta _{nl}(t,x):=\left\lbrace \begin{array}{cc}\frac{c_{2\ell }(t,x)\alpha _l(\mu _2t,\mu _2x)}{\sqrt{2e(t,x)}\gamma _{k_2}\big (Id-\frac{R_{0\ell }(t,x)}{e(t,x)}\big )}, \quad n=2\\0,\quad \quad \quad \quad n=3.\end{array}\right.\) 

Since \({\rm supp}c_{2\ell }\subseteq {\rm supp}e\) , then \(\beta _{2l}(x,t)\)  is well-defined.
Then we denote main \(l\) -perturbation \( \chi _{nol}\)  by
\(\chi _{nol}(t,x):=\left\lbrace \begin{array}{cc}\beta _{2l}(t,x)\Big (e^{i\lambda _2 2^{[l]} k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}+e^{-i\lambda _2 2^{[l]} k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}\Big ), \quad n=2\\0,\quad \quad \quad \quad n=3.\end{array}\right.\nonumber \) 

and \(l\) -correction \(\chi _{ncl}\)  by
\(\chi _{ncl}(t,x):=\left\lbrace \begin{array}{cc}\triangle \beta _{2l}(t,x)\Big (\frac{e^{i\lambda _2 2^{[l]} k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}}{-\lambda _2^22^{2[l]}|k_2|^2}+\frac{e^{-i\lambda _2 2^{[l]} k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}}{-\lambda _2^22^{2[l]}|k_2|^2}\Big )+2\nabla \beta _{2l}(t,x)\cdot \nonumber \\\nabla \Big (\frac{e^{i\lambda _2 2^{[l]} k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}}{-\lambda _2^22^{2[l]}|k_2|^2}+\frac{e^{-i\lambda _2 2^{[l]} k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}}{-\lambda _2^22^{2[l]}|k_2|^2}\Big ),\quad n=2\\0,\quad \quad \quad \quad n= 3.\end{array}\right.\nonumber \) 

Finally, we introduce \(\chi _{nl}\)  by
\(\chi _{nl}:=\chi _{nol}+\chi _{ncl}=\left\lbrace \begin{array}{cc}\triangle \Big (\beta _{2l}\Big (\frac{e^{i\lambda _2 2^{[l]} k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}}{-\lambda _2^22^{2[l]}|k_2|^2}+\frac{e^{-i\lambda _2 2^{[l]} k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}}{-\lambda _2^22^{2[l]}|k_2|^2}\Big )\Big ),\quad n=2\\0,\quad \quad \quad \quad n=3.\end{array}\right.\nonumber \) 

and \(\chi _{no}, \chi _{nc}, \chi _n\)  by, respersively,
\(\chi _{no}:=\sum _{l\in Z^3}\chi _{nol},\quad \chi _{nc}:=\sum _{l\in Z^3}\chi _{ncl},\quad \chi _n:=\sum _{l\in Z^3}\chi _{nl}.\nonumber \) 

Then \(\chi _{nol}, ~\chi _{ncl}, ~\chi _{nl}\)  and \(\chi _n\)  are all real scalar functions and as the perturbations of \(w_n\) , the summation in their definitions is finite.
Now we set
\(h_{nl}:=\left\lbrace \begin{array}{ll}\beta _{2l}-\frac{\triangle \beta _{2l}}{\lambda _2^22^{2[l]}|k_2|^2}+2\frac{\nabla \beta _{2l}\cdot k_2^{\perp }}{i\lambda 2^{[l]}|k_2|^2},~~~ n=2\\0,\quad \quad \quad \quad n=3.\end{array}\right.\quad h_{-nl}:=\left\lbrace \begin{array}{ll}\beta _{2l}-\frac{\triangle \beta _{2l}}{\lambda _2^22^{2[l]}|k_2|^2}+2\frac{\nabla \beta _{2l}\cdot k_2^{\perp }}{-i\lambda 2^{[l]}|k_2|^2},~~~ n=2\\0,\quad \quad \quad \quad n=3.\end{array}\right.\nonumber \) 

then
\(\chi _{nl}=h_{nl}e^{i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}+h_{-nl}e^{-i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}.\nonumber \) 

Moreover, since \({\rm supp} c_{2\ell }\subseteq Q_{r+\delta } \) , we know that for all \(l\in Z^3\) ,
\(\beta _{nl}\in C^{\infty }_c(Q_{r+\delta }),\quad h_{nl}\in C^{\infty }_c(Q_{r+\delta })\) 

and
\(\chi _{nol},\quad \chi _{ncl},\quad \chi _{nl},\quad \chi _{no},\quad \chi _{nc},\quad \chi _{n}\in C^{\infty }_c(Q_{r+\delta }).\nonumber \) 

Then, by (REF ), (REF ), (REF ), (REF ) and (REF ), we know that
\(\Vert \beta _{nl}\Vert _0\le \left\lbrace \begin{array}{ll}\frac{M\sqrt{\delta }}{300},~~~n=2\\0,~~~~n=3.\end{array}\right.\nonumber \) 

Therefore, by (REF )
\(\Vert \chi _{no}\Vert _0\le \left\lbrace \begin{array}{cc}\frac{M\sqrt{\delta }}{4},~~~n=2\\0,~~~~n=3.\end{array}\right.\) 

The construction of \(v_{0n}\) , \(p_{0n}\) , \(\theta _{0n}\) , \(f_{0n}\) , \(R_{0n}\)  
.
First, we denote
\(M_n\)  by
\(M_n:=&\sum \limits _{l\in Z^3}b^2_{nl}k_n\otimes k_n\Big (e^{2i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}+e^{-2i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}\Big )\nonumber \\&+\sum \limits _{l,l^{\prime }\in Z^3 ,l\ne l^{\prime }}w_{nol}\otimes w_{nol^{\prime }}\nonumber \) 

and
\(N_n,K_n\)  by
\(N_n=&\sum _{l\in Z^3}\Big [w_{nl}\otimes \Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\Big )+\Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\Big {)}\otimes w_{nl}\Big ],\nonumber \\K_n=&\sum \limits _{l\in Z^3}\beta _{nl}b_{nl}k_n\Big (e^{2i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}+e^{-2i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}\Big {)}+\sum \limits _{l,l^{\prime }\in Z^3 ,l\ne l^{\prime }}w_{nol}\chi _{nol^{\prime }}.\nonumber \) 

Then we set
\(v_{0n}:=v_{0(n-1)}+w_n,\quad p_{0n}:=&p_{0(n-1)},\quad \theta _{0n}:=\theta _{0(n-1)}+\chi _n,\nonumber \\R_{0n}:=R_{0(n-1)}+2\sum \limits _{l\in Z^3}b^2_{nl}k_n\otimes k_n+\delta R_{0n},&\quad f_{0n}:=f_{0(n-1)}+2\sum \limits _{l\in Z^3}\beta _{nl}b_{nl}k_n+\delta f_{0n},\) 

where
\(\delta R_{0n}=&\mathcal {R}(\hbox{div}M_n)+N_n-\mathcal {R}(\chi _ne_2)+\mathcal {R}\Big \lbrace \partial _tw_{n}+\hbox{div}\Big [\sum _{l\in Z^3}\Big (w_{nl}\otimes v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\nonumber \\&+v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\otimes w_{nl}\Big )\Big ]\Big \rbrace +(w_{no}\otimes w_{nc}+w_{nc}\otimes w_{no}+w_{nc}\otimes w_{nc}),\nonumber \) 

and
\(\delta f_{0n}=\left\lbrace \begin{array}{ll}\mathcal {G}(\hbox{div}K_2)+\mathcal {G}\Big (\partial _t\chi _{2}+\sum _{l\in Z^3}v_{01}\Big (\frac{l}{\mu _2}\Big )\cdot \nabla \chi _{2l}\Big )+w_{2o}\chi _{2c}+\sum _{l\in Z^3}\Big (v_{01}-v_{01}\Big (\frac{l}{\mu _2}\Big )\Big )\chi _{2l}\\+w_{2c}\chi _2+\sum _{l\in Z^3}w_{2l}\Big (\theta _{01}-\theta _{01}\Big (\frac{l}{\mu _2}\Big )\Big ),\qquad n=2\\\sum _{l\in Z^3}w_{3l}\Big (\theta _{02}-\theta _{02}\Big (\frac{l}{\mu _3}\Big )\Big ), \qquad n=3.\end{array}\right.\) 

Since
\(\hbox{div}M_n,~\chi _ne_2, ~\partial _tw_{n},~~\hbox{div}\Big [\sum _{l\in Z^3}\Big (w_{nl}\otimes v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )+v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\otimes w_{nl}\Big )\Big ]\in \Xi ,\) 
so \(\delta R_{0n}\)  is well-defined. Moreover,
\(\hbox{div}K_2,~~\partial _t\chi _{2}+\sum _{l\in Z^3}v_{01}\Big (\frac{l}{\mu _2}\Big )\cdot \nabla \chi _{2l}\in \Psi ,\) 
thus \(\delta f_{0n}\)  is well-defined.
By Proposition REF , we know that \(\delta R_{0n}\)  is a symmetric matrix and \(\delta R_{0n}\in C_c^{\infty }(Q_{r+\delta })\) . Also, by Proposition REF , we have \(\delta f_{0n}\in C_c^{\infty }(Q_{r+\delta })\) .
Obviously,
\(\hbox{div}v_{0n}=\hbox{div}v_{0(n-1)}+\hbox{div}w_{n}=0.\) 

Moreover, from the definition (REF ) on \((v_{0n}, p_{0n}, \theta _{0n}, R_{0n}, f_{0n})\)  and the fact that \((v_{0(n-1)}, p_{0(n-1)},\theta _{0(n-1)}, \\R_{0(n-1)}, f_{0(n-1)})\)  solves the system (REF ), Proposition REF , we know that
\(\begin{aligned}\hbox{div}R_{0n}=&\hbox{div}R_{0(n-1)}+\partial _tw_n-\chi _ne_2+\hbox{div}(w_{no}\otimes w_{no}+w_{n}\otimes v_{0(n-1)}+v_{0(n-1)}\otimes w_{n}\nonumber \\&+w_{no}\otimes w_{nc}+w_{nc}\otimes w_{no}+w_{nc}\otimes w_{nc})\nonumber \\=&\partial _tv_{0(n-1)}+\hbox{div}(v_{0(n-1)}\otimes v_{0(n-1)})+\nabla p_{0(n-1)}-\theta _{0(n-1)}e_2+\partial _tw_n-\chi _ne_2\nonumber \\&+\hbox{div}(w_{no}\otimes w_{no}+w_{n}\otimes v_{0(n-1)}+v_{0(n-1)}\otimes w_{n}+w_{no}\otimes w_{nc}+w_{nc}\otimes w_{no}+w_{nc}\otimes w_{nc})\nonumber \\=&\partial _tv_{0n}+\hbox{div}(v_{0n}\otimes v_{0n})+\nabla p_{0n}-\theta _{0n}e_2.\end{aligned}\) 
Where we used
\(\hbox{div}(M_n)+\hbox{div}\Big (2\sum \limits _{l\in Z^3}b^2_{nl}k_n\otimes k_n\Big )=\hbox{div}(w_{no}\otimes w_{no}).\) 
Furthermore, by (REF ) and (REF ), we have
\( \begin{aligned}f_{02}=&f_{01}+2\sum \limits _{l\in Z^3}\beta _{2l}b_{2l}k_2+\mathcal {G}(\hbox{div}K_2)+w_{2o}\chi _{2c}+\mathcal {G}\Big (\partial _t\chi _{2}+\sum _{l\in Z^3}v_{01}\Big (\frac{l}{\mu _2}\Big )\cdot \nabla \chi _{2l}\Big )\nonumber \\&+\sum _{l\in Z^3}\Big (v_0-v_{01}\Big (\frac{l}{\mu _2}\Big )\Big )\chi _{2l}+w_{2c}\chi _2+\sum _{l\in Z^3}w_{2l}\Big (\theta _{01}-\theta _{01}\Big (\frac{l}{\mu _2}\Big )\Big ).\nonumber \end{aligned}\) 
From the fact that \((v_{01}, p_{01}, \theta _{01}, R_{01}, f_{01})\)  solves the system (REF ) and Proposition REF  we have
\(\hbox{div}f_{02}=&\hbox{div}f_{01}+\partial _{t}\chi _2+\hbox{div}(w_{2o}\chi _2+w_{2c} \chi _2+v_{01} \chi _2+w_2 \theta _{01})\nonumber \\=&\hbox{div}(v_{01}\theta _{01}+w_{2o}\chi _2+w_{2c} \chi _2+v_{01} \chi _2+w_2 \theta _{01})+\partial _t(\theta _{01}+\chi _2)\nonumber \\=&\partial _t\theta _{02}+\hbox{div}(v_{02}\theta _{02}),\nonumber \) 

where we used
\(\hbox{div}K_2+\hbox{div}(2\sum \limits _{l\in Z^3}\beta _{2l}b_{2l}k_2)=\hbox{div}(w_{2o}\chi _{2o}).\nonumber \) 

Thus, the functions \((v_{02},p_{02},\theta _{02},R_{02},f_{02})\)  satisfies the system (REF ).
And from the definition (REF ) on \(\delta f_{03}\) , we have
\(\hbox{div}f_{03}=\hbox{div}f_{02}+\hbox{div}\delta f_{03}=\partial _t \theta _{02}+v_{02}\cdot \nabla \theta _{02}+w_3\cdot \nabla \theta _{02}=\partial _t \theta _{03}+v_{03}\cdot \nabla \theta _{03}.\nonumber \) 

Thus the functions \((v_{03},p_{03},\theta _{03},R_{03},f_{03})\)  also solves the system (REF ).

The n-th Representation
In this section, we will calculate the form of
\(R_{0n}+2\sum \limits _{l\in Z^3}b^2_{nl}k_n\otimes k_n={\tilde{I}}\) 
and
\(f_{0n}+2\sum \limits _{l\in Z^3}\beta _{nl}b_{nl}k_n=\tilde{II}.\) 
The term \(\tilde{I}\)
First, by (REF ), we have
\(2\sum \limits _{l\in Z^3}b^2_{nl}k_n\otimes k_n=\sum \limits _{l\in Z^3}\alpha _l^2(\mu _nt,\mu _n x)a^2_nk_n\otimes k_n=a^2_nk_n\otimes k_n.\nonumber \) 

Where we used \(\sum \limits _{l\in Z^3}\alpha _l^2=1\) .
Therefore, by (REF ), we have
\(R_{0(n-1)}+2\sum \limits _{l\in Z^3}b^2_{nl}k_n\otimes k_n=-\sum _{i=n+1}^3 a^2_ik_i\otimes k_i+\sum _{i=i}^{n-1}\delta R_{0i}.\nonumber \) 

Meanwhile, by (REF ), we have
\(R_{0n}=-\sum _{i=n+1}^3 a^2_i(t,x)k_i\otimes k_i+\sum _{i=i}^n\delta R_{0i}.\nonumber \) 

In particular,
\(R_{03}=\sum _{i=1}^3\delta R_{0i}.\nonumber \) 

In next section, we will prove that \(\delta R_{0n}\)  is small.

The term \(\tilde{II}\)
Then, by (REF ) and (REF ), we have
\(2\sum \limits _{l\in Z^3}\beta _{2l}b_{2l}k_2=\sum \limits _{l\in Z^3}\alpha _l^2(\mu _2t,\mu _2 x)c_2k_2=c_2k_2.\nonumber \) 

From the identity (REF ), we have
\(&f_{01}+2\sum \limits _{l\in Z^3}\beta _{2l}b_{2l}k_2=\delta f_{01}.\nonumber \) 

Meanwhile, by (REF ), we have
\(f_{02}=&f_{01}+2\sum \limits _{l\in Z^3}\beta _{2l}b_{2l}k_2+\delta f_{02}=\sum _{i=1}^2\delta f_{0i}.\nonumber \) 

Since \(\beta _{nl}=0\)  when \(n=3\) , then
\(f_{03}=&f_{02}+2\sum \limits _{l\in Z^3}\beta _{3l}b_{3l}k_3+\delta f_{03}=\sum _{i=1}^3\delta f_{0i}.\nonumber \) 

In next section, we will prove that \(\delta g_{0n}\)  is small.

Estimates on \(\delta R_{0n}\)  and \(\delta g_{0n}\)
We summarize the main estimates of \(b_{nl}\)  and \(\beta _{nl}\) .
Lemma 9.1 
For any \(l\in Z^3\)  and any integer \(m\ge 1\) , we have
\(\Vert b_{nl}\Vert _m+\Vert \beta _{nl}\Vert _m\le & C_m\sqrt{\delta }(\mu _n^m+\mu _n\ell ^{-(m-1)}),\\\Vert \partial _tb_{nl}\Vert _m+\Vert \partial _t\beta _{nl}\Vert _m\le & C_m\sqrt{\delta }(\mu _n^{m+1}+\mu _n\ell ^{-m}),\\\Vert \partial _{tt}b_{nl}\Vert _m+\Vert \partial _{tt}\beta _{nl}\Vert _m\le & C_m\sqrt{\delta }(\mu _n^{m+2}+\mu _n\ell ^{-m-1}),\\\Vert k_{\pm nl}\Vert _m+\Vert h_{\pm nl}\Vert _m\le & C_m\sqrt{\delta }(\mu _n^m+\mu _n\ell ^{-(m-1)}),\\\Vert \partial _tk_{\pm nl}\Vert _m+\Vert \partial _th_{\pm nl}\Vert _m\le & C_m\sqrt{\delta }(\mu _n^{m+1}+\mu _n\ell ^{-m}),\\\Vert \partial _{tt}k_{\pm nl}\Vert _m+\Vert \partial _{tt}h_{\pm nl}\Vert _m\le &C_m\sqrt{\delta }(\mu _n^{m+2}+\mu _n\ell ^{-m-1})\) 

and
\(\Vert b_{nl}\Vert _0+\Vert \beta _{nl}\Vert _0\le & C_0\sqrt{\delta },\\\Vert \partial _tb_{nl}\Vert _0+\Vert \partial _t\beta _{nl}\Vert _0\le & C_0\sqrt{\delta }\mu _n,\\\Vert \partial _{tt}b_{nl}\Vert _0+\Vert \partial _{tt}\beta _{nl}\Vert _0\le & C_0\sqrt{\delta }(\mu _n^{2}+\mu _n\ell ^{-1}),\\\Vert k_{\pm nl}\Vert _0+\Vert h_{\pm nl}\Vert _0\le & C_0\sqrt{\delta },\\\Vert \partial _tk_{\pm nl}\Vert _0+\Vert \partial _th_{\pm nl}\Vert _0\le & C_0\sqrt{\delta }\mu _n,\\\Vert \partial _{tt}k_{\pm nl}\Vert _0+\Vert \partial _{tt}h_{\pm nl}\Vert _0\le &C_0\sqrt{\delta }(\mu _n^{2}+\mu _n\ell ^{-1}).\) 
The proof is similar to lemma REF , we omit the detail here.
MM\(\square \)
Next, we give estimates on perturbations \(w_{no}, w_{nc}, \chi _{no}, \chi _{nc}\) .
Lemma 9.2 (Estimate on perturbation)
\(\Vert w_{no}\Vert _0\le &C_0\sqrt{\delta },\quad \Vert w_{no}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\lambda _n,\quad \Vert \chi _{no}\Vert _0\le C_0\sqrt{\delta },\quad \Vert \chi _{no}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\lambda _n,\nonumber \\\Vert w_{nc}\Vert _0\le & C_0\frac{\sqrt{\delta }\mu _n}{\lambda _n},\quad \Vert w_{nc}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\mu _n,\quad \Vert \chi _{nc}\Vert _0\le C_0\frac{\sqrt{\delta }\mu _n}{\lambda _n},\quad \Vert \chi _{nc}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\mu _n.\) 
The proof is similar to lemma REF , we omit the detail here.
MM\(\square \)
Corollary 9.3
\(&\Vert v_{0n}-v_{0(n-1)}\Vert _0\le \frac{M\sqrt{\delta }}{6}+C_0\frac{\sqrt{\delta }\mu _n}{\lambda _n},\quad \Vert v_{0n}-v_{0(n-1)}\Vert _{C^1_{t,x}}\le C_0\lambda _n\sqrt{\delta },\quad p_{0n}-p_{0(n-1)}=0,\nonumber \\&\Vert \theta _{02}-\theta _{01}\Vert _0\le \frac{M\sqrt{\delta }}{4}+C_0\frac{\sqrt{\delta }\mu _2}{\lambda _2},\quad \Vert \theta _{02}-\theta _{01}\Vert _{C^1_{t,x}}\le C_0\sqrt{\delta }\lambda _2,\quad \theta _{03}-\theta _{02}=0.\nonumber \) 

Estimates on \(\delta R_{0n}\)
Recalling that
\(\delta R_{0n}=&\mathcal {R}(\hbox{div}M_n)+N_n-\mathcal {R}(\chi _ne_2)+\mathcal {R}\Big \lbrace \partial _tw_{n}+\hbox{div}\Big [\sum _{l\in Z^3}\Big (w_{nl}\otimes v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\nonumber \\&+v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\otimes w_{nl}\Big )\Big ]\Big \rbrace +(w_{no}\otimes w_{nc}+w_{nc}\otimes w_{no}+w_{nc}\otimes w_{nc}).\nonumber \) 

Again, we split the stress into three parts:
(1)The oscillation part
\(\mathcal {R}(\hbox{div}M_n)-\mathcal {R}(\chi _ne_2).\) 
(2)The transport part
\(&\mathcal {R}\Big \lbrace \partial _tw_{n}+\hbox{div}\Big [\sum _{l\in Z^3}\Big (w_{nl}\otimes v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )+v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\otimes w_{nl}\Big )\Big ]\Big \rbrace \nonumber \\=&\mathcal {R}\Big (\partial _tw_{n}+\sum _{l\in Z^3}v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\cdot \nabla w_{nl}\Big ).\nonumber \) 

(3)The error part
\(N_n+(w_{no}\otimes w_{nc}+w_{nc}\otimes w_{no}+w_{nc}\otimes w_{nc}).\nonumber \) 

In the following we will estimate each term separately. Beside the estimates of \(N_n\) , the proof of other estimates are same as in section 6. We only give the proof of the estimates on \(N_n\)  and omit the others here.
Lemma 9.4 (The oscillation part)
\(\Vert \mathcal {R}({\rm div}M_n)\Vert _0\le C_0(\varepsilon )\frac{\delta \mu _n}{\lambda _n},\quad \Vert \mathcal {R}(\chi _ne_2)\Vert _0\le C_0(\varepsilon )\frac{\sqrt{\delta }}{\lambda _n},\nonumber \\\Vert \mathcal {R}({\rm div}M_n)\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\delta \mu _n,\quad \Vert \mathcal {R}(\chi _ne_2)\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sqrt{\delta }.\) 

Lemma 9.5 (The transport part)
\(\Big \Vert \mathcal {R}\Big (\partial _tw_{n}+\sum _{l\in Z^3}v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\cdot \nabla w_{nl}\Big )\Big \Vert _0\le C_0(\varepsilon )\frac{\sqrt{\delta }\mu _n}{\lambda _n},\nonumber \\\Big \Vert \mathcal {R}\Big (\partial _tw_{n}+\sum _{l\in Z^3}v_{0(n-1)}\Big (\frac{l}{\mu _{n}}\Big )\cdot \nabla w_{nl}\Big )\Big \Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sqrt{\delta }\mu _n.\) 

Lemma 9.6 (Estimate on error part I)
\(\Vert N_n\Vert _0\le C_0 \delta \frac{\lambda _{(n-1)}}{\mu _n},\qquad \Vert N_n\Vert _{C^1_{t,x}}\le C_0\lambda _n\delta \frac{\lambda _{(n-1)}}{\mu _n}.\) 
First, we have
\(N_{n}=&\sum _{l\in Z^3}\Big [w_{nl}\otimes \Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )+\Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\otimes w_{nl}\Big {)}\Big ].\nonumber \) 

By (REF ) , we have
\(&\sum _{l\in Z^3}w_{nl}\otimes \Big ( v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\nonumber \\=&\sum _{l\in Z^3}\Big (k_{nl}e^{i\lambda _n 2^{[l]} k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}+k_{-nl}e^{-i\lambda _1 2^{[l]}k_n^{\perp }\cdot \big (x-v_{0(n-1)}(\frac{l}{\mu _n})t\big )}\Big {)}\otimes \Big ( v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big ).\nonumber \) 

Obviously, \(k_{nl}(x,t)\ne 0\)  implies \(|(\mu _nt,\mu _n x)-l|\le 1\) . Moreover, by (REF ) and parameter assumption (REF ), we get \(\Vert \nabla _{t,x} v_{0(n-1)}\Vert _0\le C_0\sqrt{\delta }\lambda _{n-1}\) , therefore
\(\Big |k_{nl}\Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\Big |\le C_0\sqrt{\delta }\frac{\Vert \nabla _{t,x} v_{0(n-1)}\Vert _0}{\mu _n}\le C_0\delta \frac{\lambda _{(n-1)}}{\mu _n}.\) 
Similarly,
\(\Big |k_{-nl}\Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\Big |\le C_0\sqrt{\delta }\frac{\Vert \nabla _{t,x} v_{0(n-1)}\Vert _0}{\mu _n}\le C_0\delta \frac{\lambda _{(n-1)}}{\mu _n}.\) 
Together with (REF ), it is easy to see
\(\Big \Vert \sum _{l\in Z^3}w_{nl}\otimes \Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\Big \Vert _0\le C_0\delta \frac{\lambda _{(n-1)}}{\mu _n}.\nonumber \) 

Applying the same argument
\(\Big \Vert \sum _{l\in Z^3}\Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\otimes w_{nl}\Big \Vert _0\le C_0\delta \frac{\lambda _{(n-1)}}{\mu _n}.\nonumber \) 

Thus, we arrive at the first estimate in this lemma.
A straightforward computation gives
\(&\partial _t\Big (\sum _{l\in Z^3}w_{nl}\otimes \Big ( v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\Big )\nonumber \\=&\sum _{l\in Z^3}\partial _tw_{nl}\otimes \Big ( v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )+\sum _{l\in Z^3}w_{nl}\otimes \partial _tv_{0(n-1)}.\nonumber \) 

Thus, by (REF ), parameter assumption (REF ) and Corrollary  REF
\(\Big \Vert \partial _t\Big (\sum _{l\in Z^3}w_{nl}\otimes \Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\Big )\Big \Vert _0\le C_0\lambda _n\delta \frac{\lambda _{(n-1)}}{\mu _n}.\nonumber \) 

The same argument gives
\(\Big \Vert \nabla \Big (\sum _{l\in Z^3}w_{nl}\otimes \Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\Big )\Big \Vert _0\le C_0\lambda _n\delta \frac{\lambda _{(n-1)}}{\mu _n},\nonumber \\\Big \Vert \partial _t\Big (\sum _{l\in Z^3}\Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\otimes w_{nl}\Big )\Big \Vert _0\le C_0\lambda _n\delta \frac{\lambda _{(n-1)}}{\mu _n},\nonumber \\\Big \Vert \nabla \Big (\sum _{l\in Z^3}\Big (v_{0(n-1)}-v_{0(n-1)}\Big (\frac{l}{\mu _n}\Big )\Big )\otimes w_{nl}\Big )\Big \Vert _0\le C_0\lambda _n\delta \frac{\lambda _{(n-1)}}{\mu _n}.\nonumber \) 

Finally, collecting these estimates, we arrive at
\(\Vert N_{n}\Vert _{C^1_{t,x}}\le C_0\lambda _n\delta \frac{\lambda _{(n-1)}}{\mu _n}.\nonumber \) 

Thus, the proof of this lemma is complete.
MM\(\square \)
Lemma 9.7 (Estimates on error part II)
\(\Vert w_{no}\otimes w_{nc}+w_{nc}\otimes w_{no}+w_{nc}\otimes w_{nc}\Vert _0\le C_0\frac{\delta \mu _n}{\lambda _n},\nonumber \\\Vert w_{no}\otimes w_{nc}+w_{nc}\otimes w_{no}+w_{nc}\otimes w_{nc}\Vert _{C^1_{t,x}}\le C_0\delta \mu _n.\nonumber \) 

From Lemma REF , Lemma REF , Lemma REF  and Lemma REF , we conclude that
\(\Vert \delta R_{0n}\Vert _0\le C_0(\varepsilon )\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),\qquad \Vert \delta R_{0n}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _n\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ).\nonumber \) 
Estimate on \(\delta f_{0n}\)
We first deal with \(\delta f_{02}\) . Recalling that
\(\delta f_{02}=&\mathcal {G}(\hbox{div}K_2)+\mathcal {G}\Big (\partial _t\chi _{2}+\sum _{l\in Z^3}v_{01}\Big (\frac{l}{\mu _{2}}\Big )\cdot \nabla \chi _{2l}\Big )+w_{2o}\chi _{2c}+\sum _{l\in Z^3}\Big (v_{01}-v_{01}\Big (\frac{l}{\mu _{2}}\Big )\Big )\chi _{2l}\nonumber \\&+w_{2c}\chi _2+\sum _{l\in Z^3}w_{2l}\Big (\theta _{01}-\theta _{01}\Big (\frac{l}{\mu _2}\Big )\Big ).\nonumber \) 

As before, we split \(\delta f_{02}\)  into three parts:
(1)The oscillation part:
\(\mathcal {G}(\hbox{div}K_2).\nonumber \) 

(2)The transport part:
\(\mathcal {G}\Big (\partial _t\chi _{2}+\sum _{l\in Z^3}v_{01}\Big (\frac{l}{\mu _{2}}\Big )\cdot \nabla \chi _{2l}\Big ).\nonumber \) 

(3)The error part:
\(w_{2c}\chi _2+w_{2o}\chi _{2c}+\sum \limits _{l\in Z^3}\Big (v_{01}-v_{01}\Big (\frac{l}{\mu _{2}}\Big )\Big )\chi _{2l}+\sum _{l\in Z^3}w_{2l}\Big (\theta _{01}-\theta _{01}\Big (\frac{l}{\mu _2}\Big )\Big ).\nonumber \) 
Lemma 9.8 (The oscillation part)
\(\Vert \mathcal {G}({\rm div}K_2)\Vert _0\le C_0(\varepsilon )\frac{\delta \mu _2}{\lambda _2},\quad \Vert \mathcal {G}({\rm div}K_2)\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\delta \mu _2.\nonumber \) 

Lemma 9.9 (The transport part)
\(\Big \Vert \mathcal {G}\Big (\partial _t\chi _{2}+\sum _{l\in Z^3}v_{01}\Big (\frac{l}{\mu _{2}}\Big )\cdot \nabla \chi _{2l}\Big )\Big \Vert _0\le C_0(\varepsilon )\frac{\sqrt{\delta }\mu _2}{\lambda _2},\nonumber \\\Big \Vert \mathcal {G}\Big (\partial _t\chi _{2}+\sum _{l\in Z^3}v_{01}\Big (\frac{l}{\mu _{2}}\Big )\cdot \nabla \chi _{2l}\Big )\Big \Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sqrt{\delta }\mu _2.\nonumber \) 

Their proofs are same as in section 6.
Lemma 9.10 (The error part)
\(\Big \Vert w_{2c}\chi _2+w_{2o}\chi _{2c}+\sum _{l\in Z^3}\Big (\theta _{01}-\theta _{01}\Big (\frac{l}{\mu _{2}}\Big )\Big )\chi _{2l}\Big \Vert _0\le C_0\delta \Big (\frac{\mu _2}{\lambda _2}+\frac{\lambda _1}{\mu _2}\Big ),\nonumber \\\Big \Vert w_{2c}\chi _2+w_{2o}\chi _{2c}+\sum _{l\in Z^3}\Big (\theta _{01}-\theta _{01}\Big (\frac{l}{\mu _{2}}\Big )\Big )\chi _{2l}\Big \Vert _{C^1_{t,x}}\le C_0\lambda _2\delta \Big (\frac{\mu _2}{\lambda _2}+\frac{\lambda _1}{\mu _2}\Big ).\nonumber \) 
First, by Lemma  REF
\(\Vert w_{2c}\chi _2+w_{2o}\chi _{2c}\Vert _0\le C_0\delta \frac{\mu _2}{\lambda _2}.\nonumber \) 

As the argument in Lemma REF  , we have
\(\Big \Vert \sum _{l\in Z^3}\Big (v_{01}-v_{01}\Big (\frac{l}{\mu _{2}}\Big )\Big )\chi _{2l}\Big \Vert _0\le C_0\delta \frac{\lambda _1}{\mu _2},\qquad \Big \Vert \sum _{l\in Z^3}w_{2l}\Big (\theta _{01}-\theta _{01}\Big (\frac{l}{\mu _2}\Big )\Big )\Big \Vert _0\le C_0\delta \frac{\lambda _1}{\mu _2}.\nonumber \) 

The \(C^1_{t,x}\)  estimate is similar to that of Lemma REF .
MM\(\square \)
From the above three lemmas, we conclude
\(\Vert \delta f_{02}\Vert _0\le C_0(\varepsilon )\Big (\frac{\sqrt{\delta }\mu _2}{\lambda _2}+\delta \frac{\lambda _1}{\mu _2}\Big ),\qquad \Vert \delta f_{02}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _2\Big (\frac{\sqrt{\delta }\mu _2}{\lambda _2}+\delta \frac{\lambda _1}{\mu _2}\Big ).\nonumber \) 
Now we consider the estimates of \(\delta f_{03}\) . From definition (REF ) on \(\delta f_{03}\) , applying the same argument as in Lemma REF , we have
\(\Vert \delta f_{0n}\Vert _0\le C_0 \delta \frac{\lambda _2}{\mu _3},\qquad \Vert \delta f_{0n}\Vert _{C^1_{t,x}}\le C_0 \lambda _3\delta \frac{\lambda _2}{\mu _3}.\nonumber \) 

By inductions, we know that, for any \(2\le n \le 3\) , \((v_{0n}, p_{0n}, \theta _{0n}, R_{0n}, f_{0n})\in C_c^{\infty }(Q_{r+\delta })\)  solves system (REF ) and satisfies
\(R_{0n}=-\sum _{i=n+1}^3 a^2_ik_i\otimes k_i+\sum _{i=i}^n\delta R_{0i},\qquad g_{0n}:=\sum _{i=1}^n\delta g_{0i}\nonumber \) 

with the estimates
\(\Vert v_{0n}-v_{0(n-1)}\Vert _0\le \frac{M\sqrt{\delta }}{6}+C_0\frac{\sqrt{\delta }\mu _n}{\lambda _n} ,\quad \Vert v_{0n}-&v_{0(n-1)}\Vert _{C^1_{t,x}}\le C_0\lambda _n\sqrt{\delta },\quad \Vert p_{0n}-p_{0(n-1)}\Vert _0=0,\nonumber \\\Vert \theta _{0n}-\theta _{0(n-1)}\Vert _0\le \left\lbrace \begin{array}{cc}\frac{M\sqrt{\delta }}{4}+C_0\frac{\sqrt{\delta }\mu _2}{\lambda _2} ,~~~n=2\nonumber \\0,~~~~~~~~~~ n= 3.\end{array}\right.&\quad \Vert \theta _{0n}-\theta _{0(n-1)}\Vert _{C^1_{t,x}}\le \left\lbrace \begin{array}{cc}C_0\lambda _2\sqrt{\delta } ,~~~n=2\nonumber \\0,~~~~~~~~~~ n= 3.\end{array}\right.\\\Vert \delta R_{0n}\Vert _0\le C_0(\varepsilon )\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),&\qquad \Vert \delta f_{0n}\Vert _0\le C_0(\varepsilon )\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),\nonumber \\\Vert \delta R_{0n}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _n\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),&\qquad \Vert \delta f_{0n}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\lambda _n\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ).\nonumber \) 
Finally, we obtain \((v_{03},~p_{03},~\theta _{03},~R_{03},~f_{03})\in C_c^{\infty }(Q_{r+\delta })\)  which solves system (REF ) and satisfies
\(\Vert R_{03}\Vert _0\le C_0(\varepsilon )\sum \limits _{n=1}^{3}\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),&\quad \Vert f_{03}\Vert _0\le C_0(\varepsilon )\sum \limits _{n=1}^{3}\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),\nonumber \\\Vert R_{03}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sum \limits _{n=1}^{3}\lambda _n\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),&\qquad \Vert f_{03}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sum \limits _{n=1}^{3}\lambda _n\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),\nonumber \\\Vert v_{03}-v_0\Vert _0\le \frac{M\sqrt{\delta }}{2}+C_0\sum \limits _{n=1}^{3}\frac{\sqrt{\delta }\mu _n}{\lambda _n},&\qquad \Vert v_{03}-v_0\Vert _{C^1_{t,x}}\le C_0\sum \limits _{n=1}^{3}\lambda _n\sqrt{\delta },\nonumber \\\Vert p_{03}-p_0\Vert _0\le M\delta ,&\qquad \Vert p_{03}-p_0\Vert _{C^1_{t,x}}\le C_0, \nonumber \\\Vert \theta _{03}-\theta _0\Vert _0\le \frac{M\sqrt{\delta }}{2}+C_0\sum \limits _{n=1}^{2}\frac{\sqrt{\delta }\mu _n}{\lambda _n},&\qquad \Vert \theta _{03}-\theta _0\Vert _{C^1_{t,x}}\le C_0\sum \limits _{n=1}^{2}\lambda _n\sqrt{\delta }.\nonumber \) 
Proof of Proposition 2.1
In this section, we prove Proposition REF  by choosing the appropriate parameters
\(\ell , \mu _n, \lambda _n\)  for \(1\le n\le 3\) .
From the results of section 9, we have \((v_{03},~p_{03},~\theta _{03},~R_{03},~f_{03})\in C_c^{\infty }(Q_{r+\delta })\)  which solves system (REF ) and satisfies
\(\Vert R_{03}\Vert _0\le C_0(\varepsilon )\sum \limits _{n=1}^{3}\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),&\quad \Vert f_{03}\Vert _0\le C_0(\varepsilon )\sum \limits _{n=1}^{3}\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),\nonumber \\\Vert R_{03}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sum \limits _{n=1}^{3}\lambda _n\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),&\qquad \Vert f_{03}\Vert _{C^1_{t,x}}\le C_0(\varepsilon )\sum \limits _{n=1}^{3}\lambda _n\Big (\frac{\sqrt{\delta }\mu _n}{\lambda _n}+\delta \frac{\lambda _{n-1}}{\mu _n}\Big ),\nonumber \) 
\(\Vert v_{03}-v_0\Vert _0\le \frac{M\sqrt{\delta }}{2}+C_0\sum \limits _{n=1}^{3}\frac{\sqrt{\delta }\mu _n}{\lambda _n},&\qquad \Vert v_{03}-v_0\Vert _{C^1_{t,x}}\le C_0\sum \limits _{n=1}^{3}\lambda _n\sqrt{\delta },\nonumber \\\Vert p_{03}-p_0\Vert _0\le M\delta ,&\qquad \Vert p_{03}-p_0\Vert _{C^1_{t,x}}\le C_0, \nonumber \\\Vert \theta _{03}-\theta _0\Vert _0\le \frac{M\sqrt{\delta }}{2}+C_0\sum \limits _{n=1}^{2}\frac{\sqrt{\delta }\mu _n}{\lambda _n},&\qquad \Vert \theta _{03}-\theta _0\Vert _{C^1_{t,x}}\le C_0\sum \limits _{n=1}^{2}\lambda _n\sqrt{\delta }.\) 

where \(\lambda _0=\Lambda \delta ^{-\frac{1}{2}}+\frac{\mu _1\Lambda \ell }{\delta }\) .
We divide the remainder proof into four steps:
Step 1.
We now specify the choice of the parameters. First choose:
\(\ell =\frac{1}{L_v}\frac{\bar{\delta }}{\Lambda },\) 

with \(L_v\)  being a sufficiently large constant, which depends only on \(\Vert v\Vert _0\) .
Next, we impose
\(\mu _1=L_v\frac{\sqrt{\delta }}{\bar{\delta }}\Lambda ,\quad \lambda _1=L_v\frac{\sqrt{\delta }}{\bar{\delta }}\mu _1^{1+\varepsilon },\quad \mu _i=L_v\frac{\delta \lambda _{i-1}}{\bar{\delta }},\quad \lambda _i=L_v\frac{\sqrt{\delta }}{\bar{\delta }}\mu _i^{1+\varepsilon },\quad i=2,3.\) 

Step 2. Compatibility condition. We check that all the conditions in (REF ), (REF ) are satisfied by our choice of the parameters.
We first check the triple \((\ell , \mu _1, \lambda _1)\) . By (REF )
\(\ell ^{-1}=L_v\frac{\Lambda }{\bar{\delta }}\ge \frac{\Lambda }{\eta \delta }\) 
if we take \(L_v\ge \frac{1}{\eta }\) .
Since \(\bar{\delta }\le \delta ^{\frac{3}{2}}\) ,
\(\mu _1\ge \frac{\Lambda }{\delta }\) 
and
\(\lambda _1\ge L_v^{1+\varepsilon }\Lambda ^{1+\varepsilon }\Big (\frac{\sqrt{\delta }}{\bar{\delta }}\Big )^{2+\varepsilon }\ge \Big (L_v\frac{\Lambda }{\bar{\delta }}\Big )^{1+\varepsilon }\ge \ell ^{-(1+\varepsilon )}.\) 
It's obvious that
\(\lambda _1\ge {\mu _1^{1+\varepsilon }}.\) 
Thus, (REF ) is satisfied.
Next, for \(i=2,3\) , it's obvious that
\(\frac{\mu _i}{\mu _{i-1}}=\frac{\lambda _{i-1}}{\lambda _{i-2}}>1.\) 
A straightforward computation yield
\(\lambda _i\ge \frac{\sqrt{\delta }}{\bar{\delta }}\Big (\frac{\delta }{\bar{\delta }}\Big )^{1+\varepsilon }\lambda _{i-1}^{1+\varepsilon }\ge \lambda _{i-1}\ge \ell ^{-(1+\varepsilon )}.\) 
It's obvious that
\(\lambda _i\ge \mu _i^{1+\varepsilon }.\) 
Thus, the relationship (REF ) is satisfied.
Step 3. \(C^0\)  estimates Fixed \(\varepsilon \)  small. Thus, (REF ) implies
\(\Vert R_{03}\Vert _0\le C_0(\varepsilon ) \bar{\delta }L_v^{-1},\quad &\Vert f_{03}\Vert _0\le C_0(\varepsilon ) \bar{\delta }L_v^{-1},\nonumber \\\Vert v_{03}-v_0\Vert _0\le \frac{M\sqrt{\delta }}{2}+C_0\bar{\delta }L_v^{-1},\quad \Vert \theta _{03}-\theta _0\Vert _0&\le \frac{M\sqrt{\delta }}{2}+C_0\bar{\delta }L_v^{-1},\quad \Vert p_{03}-p_0\Vert _0\le M\delta .\nonumber \) 

Choosing \(L_v\)  sufficiently large, we can achieve the desired inequalities
(REF )-().
Step 4. \(C^1\)  estimates.
By the specified choices of parameters we have
\(\Vert R_{03}\Vert _{C^1}\le \lambda _3\bar{\delta },\quad \Vert f_{03}\Vert _{C^1}\le \lambda _3\bar{\delta },\quad \Vert v_{03}-v_0\Vert _{C^1_{t,x}}\le C_0\lambda _3\sqrt{\delta },\quad \Vert \theta _{03}-\theta _0\Vert _{C^1_{t,x}}\le C_0\lambda _3\sqrt{\delta }.\nonumber \) 

Notice that for \(i=2,3\) , there holds
\(\lambda _i=L_v^{2+\varepsilon }\frac{\sqrt{\delta }}{\bar{\delta }}\Big (\frac{\delta }{\bar{\delta }}\Big )^{1+\varepsilon }\lambda _{i-1}^{1+\varepsilon }=\frac{1}{\sqrt{\delta }}\Big (\frac{D_v\delta }{\bar{\delta }}\Big )^{2+\varepsilon }\lambda _{i-1}^{1+\varepsilon }.\) 
Thus, we conclude
\(&\max \lbrace 1,\Vert R_{03}\Vert _{C^1}, \Vert f_{03}\Vert _{C^1}, \Vert v_{03}\Vert _{C^1}, \Vert \theta _{03}\Vert _{C^1}\rbrace \le \Lambda +C_0(\varepsilon )\sqrt{\delta }\lambda _3\nonumber \\\le & C_0(\varepsilon )L_v^{(1+\varepsilon )^{2}(2+\varepsilon )+(2+\varepsilon )^2}(\sqrt{\delta })^{\varepsilon ^2+3\varepsilon +3}\Big (\frac{\sqrt{\delta }}{\bar{\delta }}\Big )^{(1+\varepsilon )^{2}(2+\varepsilon )+(2+\varepsilon )^2}\Lambda ^{(1+\varepsilon )^{3}}.\nonumber \) 

Setting \(A=C_0(\varepsilon )L_v^{(1+\varepsilon )^{2}(2+\varepsilon )+(2+\varepsilon )^2}\) , we conclude estimate (REF ).
More precisely, we have
\(\Vert \theta _{03}\Vert _{C^1_{t,x}}\le \Lambda +C_0(\varepsilon )\sqrt{\delta }\lambda _2&\le C_0(\varepsilon )L_v^{4+4\varepsilon +\varepsilon ^2}(\sqrt{\delta })^{2+\varepsilon }\Big (\frac{\sqrt{\delta }}{\bar{\delta }}\Big )^{4+4\varepsilon +\varepsilon ^2}\Lambda ^{(1+\varepsilon )^2}\nonumber \\&\le A\delta ^{\frac{2+\varepsilon }{2}}\Big (\frac{\sqrt{\delta }}{\bar{\delta }}\Big )^{4+4\varepsilon +\varepsilon ^2}\Lambda ^{(1+\varepsilon )^2},\nonumber \) 

this is the second estimate in (REF ).
Finally, we set
\(\tilde{V}:=v_{03},~\tilde{p}:=p_{03},~\tilde{\theta }:=\theta _{03},~\tilde{R}:=R_{03},~\tilde{f}:=f_{03},\) 
then \(\tilde{V},~\tilde{p},~\tilde{\theta },~\tilde{R},~\tilde{f}\)  are what we need in our Proposition (REF ).
MM\(\square \)
Proof of theorem 1.1
We first construct a non-trival solution with compact support both in space and time for system (REF ).
Construction of compactly supported solution (\(v_0, p_0, \theta _0, R_0, f_0\) ) for system (REF )
We first set \(k_1:=(1,0)^T\)  and let
\(0\le \varphi (t,x)\in C_c^{\infty }(Q_{r};R),\quad \varphi (t,x)=10M \quad {\rm in}\quad Q_{\frac{r}{2}},\) 
where \(r,M\)  is the constant appeared in Proposition (REF ). Then set
\(\bar{p}(t,x):=-2\varphi ^2(t,x),\quad a_{1l}(t,x):=\varphi (t,x)\alpha _l(\mu _1t,\mu _1x),\quad \bar{R}(t,x):=\left(\begin{array}{cc}-2\varphi ^2(t,x) & 0 \\0 & -2\varphi ^2(t,x)\end{array}\right).\nonumber \) 

Here \(\alpha _l\)  is the partition of unity in section 4. Obviously, \(\nabla \bar{p}={\rm div}\bar{R}.\)
We first set
\(v_{01ol}(t,x):=&-a_{1l}(t,x) k_1\Big (ie^{i\lambda _12^{|l|}k_1^{\perp }\cdot x}-ie^{-i\lambda _12^{|l|}k_1^{\perp }\cdot x}\Big ),\nonumber \\v_{01cl}(t,x):=&\nabla ^{\perp }(\nabla a_{1l}(t,x)\cdot k_1^{\perp })\Big (\frac{ie^{i\lambda _12^{|l|}k_1^{\perp }\cdot x}-ie^{-i\lambda _12^{|l|}k_1^{\perp }\cdot x}}{\lambda _1^22^{2|l|}}\Big )-\nabla ^{\perp }a_{1l}(t,x)\Big (\frac{e^{i\lambda _12^{|l|}k_1^{\perp }\cdot x}+e^{-i\lambda _12^{|l|}k_1^{\perp }\cdot x}}{\lambda _12^{|l|}}\Big )\nonumber \\&-\nabla a_{1l}(t,x)\cdot k_1^{\perp }\Big (\frac{k_1e^{i\lambda _12^{|l|}k_1^{\perp }\cdot x}+k_1e^{-i\lambda _12^{|l|}k_1^{\perp }\cdot x}}{\lambda _12^{|l|}}\Big ),\) 

where \(|l|\)  is the length of \(l\) , \(\mu _1\ll \lambda _1\)  are two positive numbers which will be chosen quite large, depending on appropriate norms of \(\varphi \) .
Then set
\(v_{01l}:=v_{01ol}+v_{01cl},\quad v_{01o}:=\sum \limits _{l\in Z^3}v_{01ol},\quad v_{01c}:=\sum \limits _{l\in Z^3}v_{01cl},\quad v_{01}:=\sum \limits _{l\in Z^3}v_{01l}.\nonumber \) 

Thus, a straightforward computation gives
\(v_{01}(t,x):=&\sum \limits _{l\in Z^3}\nabla ^{\perp }\hbox{div}\Big (a_{1l}(t,x)\Big (\frac{ik_1^{\perp }e^{i\lambda _12^{|l|}k_1^{\perp }\cdot x}-ik_1^{\perp }e^{-i\lambda _12^{|l|}k_1^{\perp }\cdot x}}{\lambda _1^22^{2|l|}}\Big )\Big ).\nonumber \) 

Let \(b(t,x)\in C_c^{\infty }(Q_r;R)\)  and set
\(\theta _{01o}(t,x):=&-b(t,x)(e^{i\lambda _1k_1^{\perp }\cdot x}+e^{-i\lambda _1k_1^{\perp }\cdot x}),\nonumber \\\theta _{01c}(t,x):=&\triangle b(t,x)\frac{e^{i\lambda _1k_1^{\perp }\cdot x}+e^{-i\lambda _1k_1^{\perp }\cdot x}}{\lambda _1^2}+2\nabla b(t,x)\cdot k_1^{\perp }\frac{ie^{i\lambda _1k_1^{\perp }\cdot x}-ie^{-i\lambda _1k_1^{\perp }\cdot x}}{\lambda _1}.\nonumber \) 

Then denote \(\theta _{01}\)  by
\(\theta _{01}:=&\theta _{01c}+\theta _{01o}.\nonumber \) 

Thus
\(\theta _{01}(t,x)=\triangle \Big (b(t,x)\Big (\frac{e^{i\lambda _1k_1^{\perp }\cdot x}+e^{-i\lambda _1k_1^{\perp }\cdot x}}{\lambda _1^2}\Big )\Big ).\nonumber \) 

Finally, we set
\(p_{01}:=&\bar{p},\quad R_{01}:=\bar{R}+2\sum \limits _{l\in Z^3}a^2_{1l}k_1\otimes k_1+\delta R_{01},\nonumber \\f_{01}:=&v_{01o}\theta _{01c}+v_{01c}\theta _{01o}+v_{01c}\theta _{01c}+\mathcal {G}\Big (\partial _t\theta _{01}+{\rm div}(v_{01o}\theta _{01o})\Big ),\nonumber \) 

where
\(\delta R_{01}=&v_{01o}\otimes v_{01c}+v_{01c}\otimes v_{01o}+v_{01c}\otimes v_{01c}+\mathcal {R}\Big (\partial _tv_{01}\nonumber \\&+{\rm div}\Big (-\sum \limits _{l\in Z^3}a^2_{1l}k_1\otimes k_1\Big (e^{2i\lambda _12^{|l|}k_1^{\perp }\cdot x}+e^{-i\lambda _12^{|l|}k_1^{\perp }\cdot x}\Big )+\sum \limits _{l,l^{\prime }\in Z^3,l\ne l^{\prime }}v_{01ol}\otimes v_{01ol^{\prime }}\Big )-\theta _{01}e_2\Big ).\nonumber \) 

Obviously, \({\rm div}v_{01}=0\)  and \(\partial _tv_{01},~~~ \theta _{01}e_2\in \Xi ,~~~ \partial _t\theta _{01}+{\rm div}(v_{01o}\theta _{01o})\in \Psi \) , thus \(R_{01}, f_{01}\)  is well-defined. By Proposition REF  and Proposition REF , we know that \((v_{01}, p_{01}, \theta _{01}, R_{01}, f_{01})\in C^\infty _c(Q_r)\)  solves Boussinesq-stress system (REF ). In fact, by Proposition REF , we have
\({\rm div}R_{01}=\partial _t v_{01}+{\rm div}(v_{01}\otimes v_{01})+\nabla p_{01}-\theta _{01}e_2,\nonumber \) 

where we use the identity
\(&{\rm div}\Big \lbrace -\sum \limits _{l\in Z^3}a^2_{1l}k_1\otimes k_1\Big (e^{2i\lambda _12^{|l|}k_1^{\perp }\cdot x}+e^{-i\lambda _12^{|l|}k_1^{\perp }\cdot x}\Big )+2\sum \limits _{l\in Z^3}a^2_{1l}k_1\otimes k_1+\sum \limits _{l,l^{\prime }\in Z^3,l\ne l^{\prime }}v_{01ol}\otimes v_{01ol^{\prime }}\Big \rbrace \nonumber \\=&{\rm div}(v_{01o}\otimes v_{01o}),\quad {\rm div}\bar{R}=\nabla p_{01}.\nonumber \) 

Using Proposition REF , we have
\({\rm div}f_{01}=\partial _t \theta _{01}+{\rm div}(v_{01}\theta _{01}).\nonumber \) 

Thus, \((v_{01}, p_{01}, \theta _{01}, R_{01}, f_{01})\)  solves Boussinesq-stress system (REF ).
Furthermore, we have
\(\bar{R}+2\sum \limits _{l\in Z^3}a^2_{1l}k_1\otimes k_1=\left(\begin{array}{cc}-2\varphi ^2 & 0 \\0 & -2\varphi ^2\end{array}\right)+\left(\begin{array}{cc}2\varphi ^2 & 0 \\0 & 0\end{array}\right)=\left(\begin{array}{cc}0 & 0 \\0 & -2\varphi ^2\end{array}\right),\nonumber \) 

therefore
\(R_{01}=\left(\begin{array}{cc}0 & 0 \\0 & -2\varphi ^2\end{array}\right)+\delta R_{01}.\) 

We claim \(\delta R_{01},~g_{01}\)  can be arbitrarily small by choosing appropriate \(\mu _1\)  and \(\lambda _1\) . In fact,
\(\Vert v_{01c}\Vert _0\le C_1\frac{\mu _1}{\lambda _1},\quad \Vert \theta _{01c}\Vert _0\le C_1\frac{\mu _1}{\lambda _1},\quad \Vert v_{01o}\Vert _0\le C_1,\quad \Vert \theta _{01o}\Vert _0\le C_1.\nonumber \) 

Here and sebsequent, \(C_1\)  is an absolute constant which depends on functions \(b,~~ \varphi \) . Therefore
\(\Vert v_{01o}\otimes v_{01c}+v_{01c}\otimes v_{01o}+v_{01c}\otimes v_{01c}\Vert _0\le C_1\frac{\mu _1}{\lambda _1},\quad \Vert v_{01o}\theta _{01c}+v_{01c}\theta _{01o}+v_{01c}\theta _{01c}\Vert _0\le C_1\frac{\mu _1}{\lambda _1}.\) 

Moreover,
by Proposition REF  and the same argument as REF , we have
\(&\Big \Vert \mathcal {R}\Big (\partial _tv_{01}+{\rm div}\Big (-\sum \limits _{l\in Z^3}a^2_{1l}k_1\otimes k_1\Big (e^{2i\lambda _12^{|l|}k_1^{\perp }\cdot x}+e^{-i\lambda _12^{|l|}k_1^{\perp }\cdot x}\Big )+\sum \limits _{l,l^{\prime }\in Z^3,l\ne l^{\prime }}v_{01ol}\otimes v_{01ol^{\prime }}\Big )-\theta _{01}e_2\Big )\Big \Vert _0\nonumber \\\le & C_1\frac{\mu _1}{\lambda _1}.\) 

Similarly, by Proposition REF , we obtain
\(\Big \Vert \mathcal {G}\Big (\partial _t\theta _{01}+{\rm div}(v_{01o}\theta _{01o})\Big )\Big \Vert _0\le C_1\frac{\mu _1}{\lambda _1}.\) 

Thus, combining (REF ), (REF ) and (REF ), we arrive at
\(\Vert f_{01}\Vert _0\le C_1 \frac{\mu _1}{\lambda _1},~~\Vert \delta R_{01}\Vert _0\le C_1 \frac{\mu _1}{\lambda _1}.\nonumber \) 

Hence \(\delta R_{01},~g_{01}\)  can be arbitrarily small by choosing appropriate \(\mu _1,~\lambda _1\) .
Take \(k_2:=(0,1)^T\) , \(a_{2l}(t,x):=\varphi (t,x)\alpha _l(\mu _2t,\mu _2x)\)  and set
\(w_{2ol}(t,x):=&-a_{2l}(t,x)k_2\Big (ie^{i\lambda _22^{|l|}k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}-ie^{-i\lambda _22^{|l|}k_1^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}\Big ),\nonumber \\w_{2cl}(t,x):=&\nabla ^{\perp }(\nabla a_{2l}(t,x)\cdot k_2^{\perp })\Big (\frac{ie^{i\lambda _22^{|l|}k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}-ie^{-i\lambda _22^{|l|}k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}}{\lambda _2^22^{2|l|}}\Big )\nonumber \\&-\nabla ^{\perp }a_{2l}(t,x)\Big (\frac{e^{i\lambda _22^{|l|}k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}+e^{-i\lambda _22^{|l|}k_2^{\perp }\cdot x}}{\lambda _22^{|l|}}\Big )\nonumber \\&-\nabla a_{2l}(t,x)\cdot k_2^{\perp }\Big (\frac{k_2e^{i\lambda _22^{|l|}k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}+k_2e^{-i\lambda _22^{|l|}k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}}{\lambda _22^{|l|}}\Big ),\nonumber \\w_{2l}:=&w_{2ol}+w_{2cl},\quad w_{2o}:=\sum \limits _{l\in Z^3}w_{2ol},\quad w_{2c}:=\sum \limits _{l\in Z^3}w_{2cl},\quad w_{2}:=w_{2o}+w_{2c},\) 

where \(\mu _2\ll \lambda _2\)  are two positive numbers which will be chosen quite large, depending on appropriate norms of \(v_{01}, \theta _{01}\) .
Then, a straightforward computation gives
\(w_{2}:=&\sum \limits _{l\in Z^3}\nabla ^{\perp }\hbox{div}\Big (a_{2l}\Big (\frac{ik_2^{\perp }e^{i\lambda _22^{|l|}k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}-ik_2^{\perp }e^{-i\lambda _22^{|l|}k_2^{\perp }\cdot \big (x-v_{01}(\frac{l}{\mu _2})t\big )}}{\lambda _2^22^{2|l|}}\Big )\Big ).\nonumber \) 
Finally, we set
\(v_{02}:=v_{01}+w_{2},\quad \theta _{02}:=\theta _{01},&\quad p_{02}:=p_{01}.\nonumber \\R_{02}:=R_{01}+2\sum \limits _{l\in Z^3}a^2_{2l}k_2\otimes k_2+\delta R_{02},&\quad f_{02}:=f_{01}+\delta f_{02},\nonumber \) 

where
\(\delta R_{02}=&w_{2o}\otimes w_{2c}+w_{2c}\otimes w_{2o}+w_{2c}\otimes w_{2c}+w_{2c}\otimes v_{01}+v_{01}\otimes w_{2c}+\mathcal {R}\Big (\partial _tw_{2}+{\rm div}\Big (w_{2o}\otimes v_{01}\Big (\frac{l}{\mu _2}\Big )\nonumber \\&+v_{01}\Big (\frac{l}{\mu _2}\Big )\otimes w_{2o} \Big )\Big )+{\rm div}\Big (-\sum \limits _{l\in Z^3}a^2_{2l}k_2\otimes k_2\Big (e^{2i\lambda _22^{|l|}k_2^{\perp }\cdot x}+e^{-i\lambda _22^{|l|}k_2^{\perp }\cdot x}\Big )\nonumber \\&+\sum \limits _{l,l^{\prime }\in Z^3,l\ne l^{\prime }}w_{2ol}\otimes w_{2ol^{\prime }}\Big )+w_{2o}\otimes \Big (v_{01}-v_{01}\Big (\frac{l}{\mu _2}\Big )\Big )+\Big (v_{01}-v_{01}\Big (\frac{l}{\mu _2}\Big )\Big )\otimes w_{2o},\nonumber \\\delta f_{02}=&\mathcal {G}(w_{2}\cdot \nabla \theta _{01}).\) 

Obviously, \({\rm div}v_{02}=0\)  and \(\partial _tw_{2}\in \Xi \) ,
thus by Proposition REF  and Proposition REF , \(R_{02}, f_{02}\)  are well-defined and \((v_{02}, p_{02}, \theta _{02}, R_{02}, f_{02})\in C^\infty _c(Q_r)\)  solves Boussinesq-stress system (REF ). In fact,
\({\rm div}R_{02}=&{\rm div}R_{01}+{\rm div}(w_{2o}\otimes w_{2o}+w_{2o}\otimes w_{2c}+w_{2c}\otimes w_{2o}+w_{2c}\otimes w_{2c}+w_{2o}\otimes v_{01}\nonumber \\&+v_{01}\otimes w_{2o}+w_{2c}\otimes v_{01}+v_{01}\otimes w_{2c})+\partial _tw_{2}\nonumber \\=&\partial _tv_{02}+{\rm div}(v_{02}\otimes v_{02})+\nabla p_{02}-\theta _{02}e_2,\nonumber \) 

where we use
\(&{\rm div}\Big \lbrace -\sum \limits _{l\in Z^3}a^2_{2l}k_2\otimes k_2\Big (e^{2i\lambda _22^{|l|}k_2^{\perp }\cdot x}+e^{-i\lambda _22^{|l|}k_2^{\perp }\cdot x}\Big )+2\sum \limits _{l\in Z^3}a^2_{2l}k_2\otimes k_2+\sum \limits _{l,l^{\prime }\in Z^3,l\ne l^{\prime }}w_{2ol}\otimes w_{2ol^{\prime }}\Big \rbrace \nonumber \\=&{\rm div}(w_{2o}\otimes w_{2o}).\nonumber \) 

And
\({\rm div}f_{02}={\rm div}f_{01}+{\rm div}(w_{2}\theta _{01})=\partial _t \theta _{01}+{\rm div}(v_{01}\theta _{01}+w_{2}\theta _{01})=\partial _t \theta _{02}+{\rm div}(v_{02}\theta _{02}).\nonumber \) 

We claim \(\delta R_{02},~\delta g_{02}\)  can be arbitrarily small by choosing appropriate \(\mu _2,\)  and \(\lambda _2\) . In fact,
\(\Vert w_{2c}\Vert _0\le C_2\frac{\mu _2}{\lambda _2},~~\Vert w_{2o}\Vert _0\le C_2.\nonumber \) 

Here and subsequent, \(C_2\)  is an constant which depends on appropriate norms of \(v_{01}, \theta _{01}\) . Therefore
\(\Vert w_{2o}\otimes w_{2c}+w_{2c}\otimes w_{2o}+w_{2c}\otimes w_{2c}+w_{2c}\otimes v_{01}+v_{01}\otimes w_{2c}\Vert _0\le C_2\frac{\mu _2}{\lambda _2}.\) 

By the same argument as lemma REF , Lemma REF , we obtain
\(&\Big \Vert \mathcal {R}\Big \lbrace {\rm div}\Big (-\sum \limits _{l\in Z^3}a^2_{2l}k_2\otimes k_2\Big (e^{2i\lambda _22^{|l|}k_2^{\perp }\cdot x}+e^{-i\lambda _22^{|l|}k_2^{\perp }\cdot x}\Big )+\sum \limits _{l,l^{\prime }\in Z^3,l\ne l^{\prime }}w_{2ol}\otimes w_{2ol^{\prime }}\Big )\Big \rbrace \Big \Vert _0\le C_2\frac{\mu _2}{\lambda _2},\nonumber \\&\Big \Vert \mathcal {R}\Big (\partial _tw_{2o}+v_{01}\Big (\frac{l}{\mu _1}\Big )\cdot \nabla w_{2o}\Big )\Big \Vert _0\le C_2\frac{\mu _2}{\lambda _2},\quad \Vert \mathcal {R}(\partial _tw_{2c})\Vert _0\le C_2\frac{\mu _2}{\lambda _2}.\nonumber \) 

Moreover, \(a_{2l}(t,x)\ne 0\)  implies \(|(\mu _2t,\mu _2x)-l|\le 1\) , therefore
\(\Big \Vert w_{2o}\otimes \Big (v_{01}-v_{01}\Big (\frac{l}{\mu _2}\Big )\Big )+\Big (v_{01}-v_{01}\Big (\frac{l}{\mu _2}\Big )\Big )\otimes w_{2o}\Big \Vert _0\le \frac{C_2}{\mu _2}.\nonumber \) 

Collecting the above estimates, we arrive at
\(\Vert \delta R_{02}\Vert _0\le \frac{C_2}{\mu _2}+C_2\frac{\mu _2}{\lambda _2}.\) 

By Proposition REF  and (REF ), we have
\(\Vert \delta f_{02}\Vert _0\le \frac{C_2}{\lambda _2}.\) 

Moreover, it's obvious that
\(2\sum \limits _{l\in Z^3}a^2_{2l}(t,x)k_2\otimes k_2=\left(\begin{array}{cc}0 & 0 \\0 & 2\varphi ^2(t,x)\end{array}\right).\nonumber \) 

Thus, by (REF ),
\(R_{01}+2\sum \limits _{l\in Z^3}a^2_{2l}k_2\otimes k_2=\delta R_{01}.\nonumber \) 

Finally, we have
\(R_{02}=\delta R_{01}+\delta R_{02},\quad f_{02}=f_{01}+\delta f_{02}\nonumber \) 

and
\(\Vert R_{02}\Vert _0\le \frac{C_2}{\mu _2}+ C_2\frac{\mu _2}{\lambda _2}+C_1 \frac{\mu _1}{\lambda _1},\qquad \Vert f_{02}\Vert _0\le C_1 \frac{\mu _1}{\lambda _1}+\frac{C_2}{\lambda _2}.\nonumber \) 

Next, we claim \(\Vert v_{02}\Vert _0\ge 10M\) . In fact,
\(v_{02}=v_{01}+w_2=v_{01o}+w_{2o}+v_{01c}+w_{2c}.\nonumber \) 

By (REF ) and (REF )
\(v_{01o}=2\sum \limits _{l\in Z^3}a_{1l}\Big (\sin (\lambda _12^{|l|}x_2),0\Big )^T,\quad w_{2o}=-2\sum \limits _{l\in Z^3}a_{2l}\Big (0,\sin \Big (\lambda _22^{|l|}(x_1-v^1_{01}\Big (\frac{l}{\mu _2}\Big )t\Big )\Big )^T,\nonumber \) 

where \(v^1_{01}\)  is the first component of \(v_{01}\) .
If we set \((t,x_1,x_2):=(0, \frac{\pi }{2\lambda _2}, 0)\)  and take \(1\ll \mu _2\ll \lambda _2\) , then
\(v_{01o}(t,x_1,x_2)=&0,\nonumber \\w_{2o}(t,x_1,x_2)=&-2\sum \limits _{l\in Z^3}\varphi \Big (0,\frac{\pi }{2\lambda _2},0\Big )\alpha _l\Big (0,\frac{\mu _2\pi }{2\lambda _2}, 0\Big )\Big (0,sin\Big (\frac{2^{|l|}\pi }{2}\Big )\Big )^T=(0,-20M)^T.\nonumber \) 

Moreover, we can take \(1\ll \mu _1\ll \lambda _1\ll \mu _2\ll \lambda _2\)  such that
\(\Vert v_{01c}\Vert _0+\Vert w_{2c}\Vert _0\le M.\nonumber \) 

Therefore, we conclude that
\(\Vert v_{02}\Vert _0\ge 10M.\nonumber \) 

Finally, we set \((v_0,~p_0,~\theta _0,~R_0,~f_0):=(v_{02},~p_{02},~\theta _{02},~R_{02},~f_{02})\) .
In conclusion, for any \( M>0,~ r>0\) , we can construct function \((v_0,~p_0,~\theta _0,~R_0,~f_0)\in C^{\infty }_{c}(Q_{r})\)  which solves Boussinesq-stress system (REF ) and satisfies the following estimates
\(\Vert R_{0}\Vert _0\le \frac{C_2}{\mu _2}+ C_2\frac{\mu _2}{\lambda _2}+C_1 \frac{\mu _1}{\lambda _1},\quad \Vert f_{0}\Vert _0\le C_1 \frac{\mu _1}{\lambda _1}+\frac{C_2}{\lambda _2},\quad \Vert v_{0}\Vert _0\ge 10M, \quad \theta _0\in \Theta (r,k_1).\nonumber \) 

Similarly, for \(k_2=(0,1)^T\) , we also can construct function \((v_0,~p_0,~\theta _0,~R_0,~f_0)\in C^{\infty }_{c}(Q_{r})\)  which solves Boussinesq-stress system (REF ) and satisfies the following estimates
\(\Vert R_{0}\Vert _0\le \frac{C_2}{\mu _2}+ C_2\frac{\mu _2}{\lambda _2}+C_1 \frac{\mu _1}{\lambda _1},\quad \Vert f_{0}\Vert _0\le C_1 \frac{\mu _1}{\lambda _1}+\frac{C_2}{\lambda _2},\quad \Vert v_{0}\Vert _0\ge 10M, \quad \theta _0\in \Theta (r,k_2).\nonumber \) 

Proof of Theorem 1.1
From subsection 11.1, we know that for any \(r>0\) , \(k=(0,1)^T\)  or \((1,0)^T\) , there exists \((v_0,~p_0,~\theta _0,~R_0,~f_0)\in C^{\infty }_{c}(Q_{r})\)  which solves Boussinesq-stress system (REF ) and satisfies the following estimates:
\(\Vert R_{0}\Vert _0\le \frac{C_2}{\mu _2}+ C_2\frac{\mu _2}{\lambda _2}+C_1 \frac{\mu _1}{\lambda _1},\quad \Vert f_{0}\Vert _0\le C_1 \frac{\mu _1}{\lambda _1}+\frac{C_2}{\lambda _2},\quad \Vert v_{0}\Vert _0\ge 10M,\quad \theta _0\in \Theta (r,k).\nonumber \) 

Take \(a,b \ge \frac{3}{2}\)  such that \(\frac{1}{a}\le \min \lbrace \frac{r}{2},\frac{\varepsilon ^2}{16M^2}\rbrace \)  and set \(\delta _n:=a^{-b^n}: n=0,1,2,\cdot \cdot \cdot \)  . Then taking \(\mu _1,\mu _2,\lambda _1,\lambda _2\)  with \(1\ll \mu _1\ll \lambda _1\ll \mu _2\ll \lambda _2\)  such that
\(\Vert R_{0}\Vert _0\le \eta \delta _0,~~\Vert f_{0}\Vert _0\le \eta \delta _0.\nonumber \) 

Applying Proposition REF  iteratively, we can construct
\((v_{n},~p_{n},~\theta _{n},~R_{n},~f_{n})\in C_c^\infty (Q_{r+\sum _{i=0}^n\delta _i}), \quad n=1,2,\cdot \cdot \cdot \) 
such that they solve system REF  and satisfy the following estimates
\(\Vert R_{n}\Vert _0\le &\eta \delta _n,\\\Vert f_{n}\Vert _0\le &\eta \delta _n,\\\Vert v_{n+1}-v_{n}\Vert _0 \le & M\sqrt{\delta _n},\\\Vert \theta _{n+1}-\theta _{n}\Vert _0 \le & M\sqrt{\delta _n},\\\Vert p_{n+1}-p_{n}\Vert _0 \le & M\delta _n,\\\Lambda _{n+1}:=\max \lbrace 1, \Vert R_{n+1}\Vert _{C^1_{t,x}},& \Vert f_{n+1}\Vert _{C^1_{t,x}}, \Vert v_{n+1}\Vert _{C^1_{t,x}}, \Vert \theta _{n+1}\Vert _{C^1_{t,x}}\rbrace \nonumber \\\le A(\sqrt{\delta _n})^{\varepsilon ^2+3\varepsilon +3}&\Big (\frac{\sqrt{\delta _n}}{\delta _{n+1}}\Big )^{(1+\varepsilon )^{2}(2+\varepsilon )+(2+\varepsilon )^2}\Lambda _n^{(1+\varepsilon )^{3}} .\) 

In particular,
\(\Vert p_n\Vert _{C^1_{t,x}}&\le C_0,\qquad \Vert \theta _{n+1}\Vert _{C^1_{t,x}}\le A(\sqrt{\delta _n})^{2+\varepsilon }\Big (\frac{\sqrt{\delta _n}}{\delta _{n+1}}\Big )^{4+4\varepsilon +\varepsilon ^2}\Lambda _n^{(1+\varepsilon )^2}.\) 

It is obvious that \(\sum _{i=0}^\infty \delta _i<r\) .
Thus, by (11.10)-(), we know that \((v_{n},~p_{n},~\theta _{n},~R_{n},~f_{n})\)  are Cauchy sequence in \(C_c(Q_{2r})\) , therefore there exists
\(\begin{aligned}(v,p,\theta )\in C_c(Q_{2r})\end{aligned}\) 
such that
\(\begin{aligned}v_{n}\rightarrow v,\quad p_{n}\rightarrow p,\quad \theta _{n}\rightarrow \theta ,\quad R_{n}\rightarrow 0,\quad f_{n}\rightarrow 0\end{aligned}\) 
in \(C_c(Q_{2r})\) .
By (REF )
\(\Vert v-v_0\Vert _0\le M\sum \limits _{n=0}^{\infty }\sqrt{\delta _n}<4M.\) 
Therefore
\(\Vert v\Vert _0\ge 6M,~~v\ne 0.\) 
By ()
\(\Vert \theta -\theta _0\Vert _0\le M\sum \limits _{n=0}^{\infty }\sqrt{\delta _n}\le \frac{4M}{\sqrt{a}}<\varepsilon .\) 
Passing into the limit in (REF ), we conclude that \(v,~p,~\theta \)  solve (REF ) in the sense of distribution.
Next, we prove that the solution \(v , p, \theta \)  is Hölder continuous. We claim that for a suitable choice of \(a, b\) , there exist a constant \(c> 1\)  such that
\(\Lambda _n\le a^{cb^n}.\) 
We prove this claim by induction.
Indeed, for \(n=0\) , it's obvious if we take \(a\ge \Lambda _0:=\max \lbrace 1, \Vert R_0\Vert _0, \Vert f_0\Vert _0, \Vert v_0\Vert _0, \Vert \theta _0\Vert _0\rbrace \) . Assuming that we have proved \(\Lambda _n\le a^{cb^n}\) , then
\(\Lambda _{n+1}\le A(\sqrt{\delta _n})^{\varepsilon ^2+3\varepsilon +3}\Big (\frac{\sqrt{\delta _n}}{\delta _{n+1}}\Big )^{(1+\varepsilon )^{2}(2+\varepsilon )+(2+\varepsilon )^2}\Lambda _n^{(1+\varepsilon )^{3}}\le Aa^{-\frac{\varepsilon }{2}b^n}a^{cb^{n+1}}a^{\big ((b-\frac{1}{2})d-\frac{\varepsilon ^2+3\varepsilon +3}{2}+\frac{\varepsilon }{2}+c(1+\varepsilon )^3-cb\big )b^n}.\nonumber \) 

where \(d=(1+\varepsilon )^{2}(2+\varepsilon )+(2+\varepsilon )^2\) .
Take \(b=\frac{3}{2}\)  and \(c=\frac{2d-(\varepsilon ^2+2\varepsilon +3)}{3-2(1+\varepsilon )^3}\)  , we arrive at
\(\Lambda _{n+1}\le Aa^{-\frac{\varepsilon }{2}}a^{cb^{n+1}}.\) 
Then choosing \(a\ge A^{\frac{2}{\varepsilon }}\) , we have \(\Lambda _{n+1}\le a^{cb^{n+1}}.\)  Finally, we take
\(a:=\max \Big \lbrace A^{\frac{2}{\varepsilon }},~~ \Lambda _0,~~~ \frac{3}{2},~~~\frac{1}{\min \lbrace \frac{r}{2},\frac{\varepsilon ^2}{16M^2}\rbrace }\Big \rbrace ,\) 
then \(a\) 
satisfies all the needed conditions.
Now we consider the approximate sequence \(v_n, p_n, \theta _n\) . By (REF ), we have
\(\Vert v_{n+1}-v_n\Vert _0\le Ma^{-\frac{1}{2}b^n}.\nonumber \) 

Moreover, we have
\(\Vert v_{n+1}-v_n\Vert _{C^1_{t,x}}\le \Lambda _n+\Lambda _{n+1}\le 2a^{cb^{n+1}}.\nonumber \) 

Therefore, for any \(\alpha \in (0,1)\)
\(\Vert v_{n+1}-v_n\Vert _{C^\alpha _{t,x}}\le 2Ma^{\big (\alpha cb-\frac{(1-\alpha )}{2}\big )b^n}.\nonumber \) 

If \(\alpha <\frac{1}{1+2bc}\) , then \(\alpha cb-\frac{(1-\alpha )}{2}< 0\) , thus \(v_n\)  are Cauchy sequence in \(C^\alpha _{t,x}.\)  Take the value of \(b,c\) , we kwon that \(v\in C^\alpha _{t,x}\)  for any \(\alpha <\frac{3-2(1+\varepsilon )^3}{3-2(1+\varepsilon )^3+6d-3(\varepsilon ^2+2\varepsilon +3)}\) . When \(\varepsilon \rightarrow 0\) , we have
\(\frac{3-2(1+\varepsilon )^3}{3-2(1+\varepsilon )^3+6d-3(\varepsilon ^2+2\varepsilon +3)}\rightarrow \frac{1}{28}\) .
By () and (REF ), we know that \(p\in C^\beta _{t,x}\)  for any \(\beta \in (0,1)\) .
By () and (REF ), we have
\(\Vert \theta _{n+1}-\theta _n\Vert _0\le Ma^{-\frac{1}{2}b^n}\nonumber \) 

and
\(\Vert \theta _{n+1}-\theta _n\Vert _{C^1_{t,x}}\le 2a^{(3+4\varepsilon +\varepsilon ^2+c(1+\varepsilon )^2)b^n}.\nonumber \) 

By interpolation, for any \(\gamma \in (0,1)\) , we have
\(\Vert \theta _{n+1}-\theta _n\Vert _{C^\gamma _{t,x}}\le 2Ma^{\big (\gamma (3+4\varepsilon +\varepsilon ^2+c(1+\varepsilon )^2)-\frac{1-\gamma }{2}\big )b^n}.\nonumber \) 

Take \(\gamma <\frac{1}{1+2(3+4\varepsilon +\varepsilon ^2+c(1+\varepsilon )^2)}\) , then \(\theta _n\)  converge in \(C^\gamma \) , which implies that \(\theta \in C^\gamma _{t, x}\)  for any \(\gamma <\frac{1}{1+2(3+4\varepsilon +\varepsilon ^2+c(1+\varepsilon )^2)}\) . When \(\varepsilon \rightarrow 0\) , we have \(\frac{1}{1+2(3+4\varepsilon +\varepsilon ^2+c(1+\varepsilon )^2)}\rightarrow \frac{1}{25}\) .
Thus, we complete our proof of theorem REF .
MM\(\square \)
Acknowledgments.
The research is partially supported by the Chinese NSF under grant 11471320.
We thank Tianwen Luo for very valuable discussions.
